103 trigonometry problems

About the Authors

Titu Andreescu received his BA, MS, and PhD from the West Universityof Timisoara, Romania. The topic of his doctoral dissertation was "Researchon Diophantine Analysis and Applications". Titu served as director of theMAA American Mathematics Competitions (1998-2003), coach of the USAInternational Mathematical Olympiad Team (IMO) for 10 years (1993-2002), director of the Mathematical Olympiad Summer Program (1995-2002) and leader of the USA IMO Team (1995-2002). In 2002 Titu waselected member of the IMO Advisory Board, the governing body of theinternational competition. Titu received the Edyth May Sliffe Award forDistinguished High School Mathematics Teaching from the MAA in 1994and a "Certificate of Appreciation" from the president of the MAA in 1995for his outstanding service as coach of the Mathematical Olympiad SummerProgram in preparing the US team for its perfect performance in Hong Kongat the 1994 International Mathematical Olympiad.

Zuming Feng graduated with a PhD from Johns Hopkins University withemphasis on Algebraic Number Theory and Elliptic Curves. He teaches atPhillips Exeter Academy. He also served as a coach of the USA IMO team(1997-2003), the deputy leader of the USA IMO Team (2000-2002), and anassistant director of the USA Mathematical Olympiad Summer Program(1999-2002). He is a member of the USA Mathematical Olympiad Commit-tee since 1999, and has been the leader of the USA IMO team and theacademic director of the USA Mathematical Olympiad Summer Programsince 2003. He received the Edyth May Sliffe Award for Distinguished HighSchool Mathematics Teaching from the MAA in 1996 and 2002.

Titu Andreescu

Zuming Feng

103 Trigonometry ProblemsFrom the Training of the USA IMO Team

BirkhäuserBoston • Basel • Berlin

Titu AndreescuUniversity of WisconsinDepartment of Mathematical and Computer SciencesWhitewater, WI 53190U.S.A.

Library of Congress Cataloging-in-Publication DataAndreescu, Titu, 1956- 103 trigonometry problems : from the training of the USA IMO team / Titu Andreescu, Zuming Feng. p. cm. Includes bibliographical references. ISBN 0-8176-4334-6 (acid-free paper) 1. Trigonometry–Problems, exercises, etc. I. Title: One hundred and three trigonometry problems. II. Feng, Zuming. III. Title.

QA537.A63 2004 516.24–dc22 2004045073

©2005 Birkhäuser Boston®

Birkhäuser

AMS Subject Classifications: Primary: 97U40, 00A05, 00A07, 51-XX; Secondary: 11L03, 26D05,33B10, 42A05

All rights reserved. This work may not be translated or copied in whole or in part without the writtenpermission of the publisher (Birkhäuser Boston, c/o Springer Science+Business Media Inc., Rightsand Permissions, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts inconnection with reviews or scholarly analysis. Use in connection with any form of informationstorage and retrieval, electronic adaptation, computer software, or by similar or dissimilar method-ology now known or hereafter developed is forbidden.The use in this publication of trade names, trademarks, service marks and similar terms, even if theyare not identified as such, is not to be taken as an expression of opinion as to whether or not they aresubject to proprietary rights.

Printed in the United States of America.

9 8 7 6 5 4 3 2 1 SPIN 10982723

www.birkhauser.com

Zuming FengPhillips Exeter AcademyDepartment of MathematicsExeter, NH 03833U.S.A.

ISBN 0-8176-4334-6 Printed on acid-free paper.

Contents

Preface vii

Acknowledgments ix

Abbreviations and Notation xi

1 Trigonometric Fundamentals 1Definitions of Trigonometric Functions in Terms of Right Triangles 1Think Within the Box 4You’ve Got the Right Angle 6Think Along the Unit Circle 10Graphs of Trigonometric Functions 14The Extended Law of Sines 18Area and Ptolemy’s Theorem 19Existence, Uniqueness, and Trigonometric Substitutions 23Ceva’s Theorem 28Think Outside the Box 33Menelaus’s Theorem 33The Law of Cosines 34Stewart’s Theorem 35Heron’s Formula and Brahmagupta’s Formula 37Brocard Points 39

vi Contents

Vectors 41The Dot Product and the Vector Form of the Law of Cosines 46The Cauchy–Schwarz Inequality 47Radians and an Important Limit 47Constructing Sinusoidal Curves with a Straightedge 50Three Dimensional Coordinate Systems 51Traveling on Earth 55Where Are You? 57De Moivre’s Formula 58

2 Introductory Problems 63

3 Advanced Problems 73

4 Solutions to Introductory Problems 83

5 Solutions to Advanced Problems 125

Glossary 199

Further Reading 211

Preface

This book contains 103 highly selected problems used in the training and testing ofthe U.S. International Mathematical Olympiad (IMO) team. It is not a collection ofvery difficult, impenetrable questions. Instead, the book gradually builds students’trigonometric skills and techniques. The first chapter provides a comprehensive in-troduction to trigonometric functions, their relations and functional properties, andtheir applications in the Euclidean plane and solid geometry. This chapter can serveas a textbook for a course in trigonometry. This work aims to broaden students’view of mathematics and better prepare them for possible participation in variousmathematical competitions. It provides in-depth enrichment in important areas oftrigonometry by reorganizing and enhancing problem-solving tactics and strategies.The book further stimulates interest for the future study of mathematics.

In the United States ofAmerica, the selection process leading to participation in theInternational Mathematical Olympiad (IMO) consists of a series of national contestscalled the American Mathematics Contest 10 (AMC 10), the American MathematicsContest 12 (AMC 12), theAmerican Invitational Mathematics Examination (AIME),and the United States of America Mathematical Olympiad (USAMO). Participationin the AIME and the USAMO is by invitation only, based on performance in thepreceding exams of the sequence. The Mathematical Olympiad Summer Program(MOSP) is a four-week intensive training program for approximately 50 very promis-ing students who have risen to the top in the American Mathematics Competitions.The six students representing the United States ofAmerica in the IMO are selected onthe basis of their USAMO scores and further testing that takes place during MOSP.

viii Preface

Throughout MOSP, full days of classes and extensive problem sets give studentsthorough preparation in several important areas of mathematics. These topics in-clude combinatorial arguments and identities, generating functions, graph theory,recursive relations, sums and products, probability, number theory, polynomials,functional equations, complex numbers in geometry, algorithmic proofs, combina-torial and advanced geometry, functional equations, and classical inequalities.

Olympiad-style exams consist of several challenging essay problems. Correctsolutions often require deep analysis and careful argument. Olympiad questions canseem impenetrable to the novice, yet most can be solved with elementary high schoolmathematics techniques, cleverly applied.

Here is some advice for students who attempt the problems that follow.

• Take your time! Very few contestants can solve all the given problems.

• Try to make connections between problems. An important theme of this workis that all important techniques and ideas featured in the book appear morethan once!

• Olympiad problems don’t “crack” immediately. Be patient. Try different ap-proaches. Experiment with simple cases. In some cases, working backwardsfrom the desired result is helpful.

• Even if you can solve a problem, do read the solutions. They may containsome ideas that did not occur in your solutions, and they may discuss strategicand tactical approaches that can be used elsewhere. The solutions are alsomodels of elegant presentation that you should emulate, but they often obscurethe tortuous process of investigation, false starts, inspiration, and attention todetail that led to them. When you read the solutions, try to reconstruct thethinking that went into them. Ask yourself, “What were the key ideas? Howcan I apply these ideas further?”

• Go back to the original problem later, and see whether you can solve it in adifferent way. Many of the problems have multiple solutions, but not all areoutlined here.

• Meaningful problem-solving takes practice. Don’t get discouraged if you havetrouble at first. For additional practice, use the books on the reading list.

Acknowledgments

Thanks to Dorin Andrica and Avanti Athreya, who helped proofread the originalmanuscript. Dorin provided acute mathematical ideas that improved the flavor ofthis book, while Avanti made important contributions to the final structure of thebook. Thanks to David Kramer, who copyedited the second draft. He made a numberof corrections and improvements. Thanks to Po-Ling Loh, Yingyu Gao, and KenneHon, who helped proofread the later versions of the manuscript.

Many of the ideas of the first chapter are inspired by the Math 2 and Math 3 teachingmaterials from the Phillips Exeter Academy. We give our deepest appreciation to theauthors of the materials, especially to Richard Parris and Szczesny “Jerzy” Kaminski.

Many problems are either inspired by or adapted from mathematical contests indifferent countries and from the following journals:

• High-School Mathematics, China

• Revista Matematica Timisoara, Romania

We did our best to cite all the original sources of the problems in the solution sec-tion. We express our deepest appreciation to the original proposers of the problems.

Abbreviations and Notation

Abbreviations

AHSME American High School Mathematics ExaminationAIME American Invitational Mathematics ExaminationAMC10 American Mathematics Contest 10AMC12 American Mathematics Contest 12,

which replaces AHSMEAPMC Austrian–Polish Mathematics CompetitionARML American Regional Mathematics LeagueIMO International Mathematical OlympiadUSAMO United States of America Mathematical OlympiadMOSP Mathematical Olympiad Summer ProgramPutnam The William Lowell Putnam Mathematical CompetitionSt. Petersburg St. Petersburg (Leningrad) Mathematical Olympiad

xii Abbreviations and Notation

Notation for Numerical Sets and Fields

Z the set of integersZn the set of integers modulo n

N the set of positive integersN0 the set of nonnegative integersQ the set of rational numbersQ+ the set of positive rational numbersQ0 the set of nonnegative rational numbersQn the set of n-tuples of rational numbersR the set of real numbersR+ the set of positive real numbersR0 the set of nonnegative real numbersRn the set of n-tuples of real numbersC the set of complex numbers[xn](p(x)) the coefficient of the term xn in the polynomial p(x)

Notation for Sets, Logic, and Geometry

|A| the number of elements in the set A

A ⊂ B A is a proper subset of B

A ⊆ B A is a subset of B

A \ B A without B (set difference)A ∩ B the intersection of sets A and B

A ∪ B the union of sets A and B

a ∈ A the element a belongs to the set A

a, b, c lengths of sides BC, CA, AB of triangle ABC

A, B, C angles � CAB, � ABC, � BCA of triangle ABC

R, r circumradius and inradius of triangle ABC

[F] area of region F[ABC] area of triangle ABC

|BC| length of line segment BC

AB the arc of a circle between points A and B

103 Trigonometry Problems

1Trigonometric Fundamentals

Definitions of Trigonometric Functions in Terms of RightTriangles

Let S and T be two sets. A function (or mapping or map) f from S to T (writtenas f : S → T ) assigns to each s ∈ S exactly one element t ∈ T (written f (s) = t);t is the image of s. For S′ ⊆ S, let f (S′) (the image of S′) denote the set of imagesof s ∈ S′ under f . The set S is called the domain of f , and f (S) is the range of f .

For an angle θ (Greek “theta") between 0◦ and 90◦, we define trigonometricfunctions to describe the size of the angle. Let rays OA and OB form angle θ (seeFigure 1.1). Choose point P on ray OA. Let Q be the foot (that is, the bottom) ofthe perpendicular line segment from P to the ray OB. Then we define the sine (sin),cosine (cos), tangent (tan), cotangent (cot), cosecant (csc), and secant (sec) functionsas follows, where |PQ| denotes the length of the line segment PQ:

sin θ = |PQ||OP | , csc θ = |OP |

|PQ| ,

cos θ = |OQ||OP | , sec θ = |OP |

|OQ| ,

tan θ = |PQ||OQ| , cot θ = |OQ|

|PQ| .

2 103 Trigonometry Problems

First we need to show that these functions are well defined; that is, they only dependson the size of θ , but not the choice of P . Let P1 be another point lying on ray OA,and let Q1 be the foot of perpendicular from P1 to ray OB. (By the way, “P sub1" is how P1 is usually read.) Then it is clear that right triangles OPQ and OP1Q1are similar, and hence pairs of corresponding ratios, such as |PQ|

|OP | and |P1Q1||OP1| , are allequal. Therefore, all the trigonometric functions are indeed well defined.

A

BO

P

Q

P1

Q1

Figure 1.1.

By the above definitions, it is not difficult to see that sin θ , cos θ , and tan θ arethe reciprocals of csc θ, sec θ , and cot θ , respectively. Hence for most purposes, it isenough to consider sin θ , cos θ , and tan θ . It is also not difficult to see that

sin θ

cos θ= tan θ and

cos θ

sin θ= cot θ.

By convention, in triangle ABC, we let a, b, c denote the lengths of sides BC, CA,and AB, and let � A, � B, and � C denote the angles CAB, ABC, and BCA. Now,consider a right triangle ABC with � C = 90◦ (Figure 1.2).

A

B Ca

b

c

Figure 1.2.

For abbreviation, we write sin A for sin � A. We have

sin A = a

c, cos A = b

c, tan A = a

b;

sin B = b

c, cos B = a

c, tan B = b

a;

1. Trigonometric Fundamentals 3

anda = c sin A, a = c cos B, a = b tan A;b = c sin B, b = c cos A, b = a tan B;c = a csc A, c = a sec B, c = b csc B, c = b sec A.

It is then not difficult to see that if A and B are two angles with 0◦ < A, B < 90◦ andA+B = 90◦, then sin A = cos B, cos A = sin B, tan A = cot B, and cot A = tan B.In the right triangle ABC, we have a2 + b2 = c2. It follows that

(sin A)2 + (cos A)2 = a2

c2 + b2

c2 = 1.

It can be confusing to write (sin A)2 as sin A2. (Why?) For abbreviation, we write(sin A)2 as sin2 A. We have shown that for 0◦ < A < 90◦,

sin2 A + cos2 A = 1.

Dividing both sides of the above equation by sin2 A gives

1 + cot2 A = csc2 A, or csc2 A − cot2 A = 1.

Similarly, we can obtain

tan2 A + 1 = sec2 A, or sec2 A − tan2 A = 1.

Now we consider a few special angles.In triangle ABC, suppose � A = � B = 45◦, and hence |AC| = |BC| (Figure 1.3,

left). Then c2 = a2 +b2 = 2a2, and so sin 45◦ = sin A = ac

= 1√2

=√

22 . Likewise,

we have cos 45◦ =√

22 and tan 45◦ = cot 45◦ = 1.

A

B

C DA

B

CFigure 1.3.

In triangle ABC, suppose � A = 60◦ and � B = 30◦ (Figure 1.3, right). Wereflect A across line BC to point D. By symmetry, � D = 60◦, so triangle ABD

is equilateral. Hence, |AD| = |AB| and |AC| = |AD|2 . Because ABC is a right


triangle, |AB|2 = |AC|2 + |BC|2. So we have |BC|2 = |AB|2 − |AB|24 = 3|AB|2

4 ,

or |BC| =√

3|AB|2 . It follows that sin 60◦ = cos 30◦ =

√3

2 , sin 30◦ = cos 60◦ = 12 ,

tan 30◦ = cot 60◦ =√

33 , and tan 60◦ = cot 30◦ = √

3.We provide one exercise for the reader to practice with right-triangle trigonometric

functions. In triangle ABC (see Figure 1.4), � BCA = 90◦, and D is the foot ofthe perpendicular line segment from C to segment AB. Given that |AB| = x and� A = θ , express all the lengths of the segments in Figure 1.4 in terms of x and θ .

A

BC

D

Figure 1.4.

Think Within the Box

For two angles α (Greek “alpha") and β (Greek “beta") with 0◦ < α, β, α + β <

90◦, it is not difficult to note that the trigonometric functions do not satisfy theadditive distributive law; that is, identities such as sin(α + β) = sin α + sin β andcos(α+β) = cos α+cos β are not true. For example, setting α = β = 30◦, we havecos(α + β) = cos 60◦ = 1

2 , which is not equal to cos α + cos β = 2 cos 30◦ = √3.

Naturally, we might ask ourselves questions such as how sin α, sin β, and sin(α +β)

relate to one another.Consider the diagram of Figure 1.5. Let DEF be a right triangle with � DEF =

90◦, � FDE = β, and |DF | = 1 inscribed in the rectangle ABCD. (This canalways be done in the following way. Construct line �1 passing through D outsideof triangle DEF such that lines �1 and DE form an acute angle congruent to α.Construct line �2 passing through D and perpendicular to line �1. Then A is the footof the perpendicular from E to line �1, and C the foot of the perpendicular from F

to �2. Point B is the intersection of lines AE and CF .)


A B

CD

E

F

�

�

��

�

Figure 1.5.

We compute the lengths of the segments inside this rectangle. In triangle DEF ,we have |DE| = |DF | · cos β = cos β and |EF | = |DF | · sin β = sin β. In triangleADE, |AD| = |DE| · cos α = cos α cos β and |AE| = |DE| · sin α = sin α cos β.Because � DEF = 90◦, it follows that � AED+� BEF = 90◦ = � AED+� ADE,and so � BEF = � ADE = α. (Alternatively, one may observe that right trianglesADE and BEF are similar to each other.) In triangle BEF , we have |BE| =|EF | · cos α = cos α sin β and |BF | = |EF | · sin α = sin α sin β. Since AD ‖ BC,� DFC = � ADF = α + β. In right triangle CDF , |CD| = |DF | · sin(α + β) =sin(α + β) and |CF | = |DF | · cos(α + β) = cos(α + β).

From the above, we conclude that

cos α cos β = |AD| = |BC| = |BF | + |FC| = sin α sin β + cos(α + β),

implying thatcos(α + β) = cos α cos β − sin α sin β.

Similarly, we have

sin(α + β) = |CD| = |AB| = |AE| + |EB| = sin α cos β + cos α sin β;that is,

sin(α + β) = sin α cos β + cos α sin β.

By the definition of the tangent function, we obtain

tan(α + β) = sin(α + β)

cos(α + β)= sin α cos β + cos α sin β

cos α cos β − sin α sin β

=sin αcos α

+ sin βcos β

1 − sin α sin βcos α cos β

= tan α + tan β

1 − tan α tan β.

We have thus proven the addition formulas for the sine, cosine, and tangent functionsfor angles in a restricted interval. In a similar way, we can develop an addition formulafor the cotangent function. We leave it as an exercise.


By setting α = β in the addition formulas, we obtain the double-angle formulas

sin 2α = 2 sin α cos α, cos 2α = cos2 α − sin2 α, tan 2α = 2 tan α

1 − tan2 α,

where for abbreviation, we write sin(2α) as sin 2α. Setting β = 2α in the additionformulas then gives us the triple-angle formulas. We encourage the reader to deriveall the various forms of the double-angle and triple-angle formulas listed in theGlossary of this book.

You’ve Got the Right Angle

Because of the definitions of the trigonometric functions, it is more convenient todeal with trigonometric functions in the context of right triangles. Here are threeexamples.

A B

C

DE

�

�

Figure 1.6.

Example 1.1. Figure 1.6 shows a long rectangular strip of paper, one corner of whichhas been folded over along AC to meet the opposite edge, thereby creating angleθ (� CAB in Figure 1.6). Given that the width of the strip is w inches, express thelength of the crease AC in terms of w and θ . (We assume that θ is between 0◦ and45◦, so the real folding situation is consistent with the configuration shown in Figure1.6.)

We present two solutions.

First Solution: In the right triangle ABC, we have |BC| = |AC| sin θ . In the righttriangle AEC, we have |CE| = |AC| sin θ . (Indeed, by folding, triangles ABC

and AEC are congruent.) Because � BCA = � ECA = 90◦ − θ , it follows that� BCE = 180◦ − 2θ and � DCE = 2θ (Figure 1.7). Then, in the right triangleCDE, |CD| = |CE| cos 2θ . Putting the above together, we have

w = |BD| = |BC| + |CD| = |AC| sin θ + |AC| sin θ cos 2θ,


implying that

|AC| = w

sin θ(1 + cos 2θ).

A B

C

DEF

�

�

��

��

Figure 1.7.

Second Solution: Let F be the foot of the perpendicular line segment from A tothe opposite edge. Then in the right triangle AEF , � AEF = 2θ and |AF | = w.Thus |AF | = |AE| sin 2θ , or |AE| = w

sin 2θ. In the right triangle AEC, � CAE =

� CAB = θ and |AE| = |AC| cos θ . Consequently,

|AC| = |AE|cos θ

= w

sin 2θ cos θ.

Putting these two approaches together, we have

|AC| = w

sin θ(1 + cos 2θ)= w

sin 2θ cos θ,

or sin θ(1 + cos 2θ) = sin 2θ cos θ . Interested readers can use the formulas wedeveloped earlier to prove this identity.

Example 1.2. In the trapezoid ABCD (Figure 1.8), AB ‖ CD, |AB| = 4 and|CD| = 10. Suppose that lines AC and BD intersect at right angles, and that linesBC and DA, when extended to point Q, form an angle of 45◦. Compute [ABCD],the area of trapezoid ABCD.


A B

C

P

Q

D

Figure 1.8.

Solution: Let segments AC and BD meet at P . Because AB ‖ CD, trianglesABP and CDP are similar with a side ratio of |AB|

|CD| = 25 . Set |AP | = 2x and

|BP | = 2y. Then |CP | = 5x and |DP | = 5y. Because � APB = 90◦, [ABCD] =12 |AC| · |BD| = 49xy

2 . (To see this, consider the following calculation: [ABCD] =[ABD] + [CBD] = 1

2 |AP | · |BD| + 12 |CP | · |BD| = 1

2 |AC| · |BD|.)Let α = � ADP and β = � BCP . In right triangles ADP and BCP , we have

tan α = |AP ||DP | = 2x

5yand tan β = |BP |

|CP | = 2y

5x.

Note that � CPD = � CQD + � QCP + � QDP , implying that α + β = � QCP +� QDP = 45◦. By the addition formulas, we obtain that

1 = tan 45◦ = tan(α + β) = tan α + tan β

1 − tan α tan β=

2x5y

+ 2y5x

1 − 2x5y

2y5x

= 10(x2 + y2)

21xy,

which establishes that xy = 10(x2+y2)21 . In triangle ABP , we have |AB|2 = |AP |2 +

|BP |2, or 16 = 4(x2 + y2). Hence x2 + y2 = 4, and so xy = 4021 . Consequently,

[ABCD] = 49xy

2= 49

2· 40

21= 140

3.

Example 1.3. [AMC12 2004] In triangle ABC, |AB| = |AC| (Figure 1.9). PointsD and E lie on ray BC such that |BD| = |DC| and |BE| > |CE|. Supposethat tan � EAC, tan � EAD, and tan � EAB form a geometric progression, and thatcot � DAE, cot � CAE, and cot � DAB form an arithmetic progression. If |AE| =10, compute [ABC], the area of triangle ABC.


A

B CD E

Figure 1.9.

Solution: We consider right triangles ABD, ACD, and ADE. Set α = � EAD

and β = � BAD = � DAC. Then � EAC = α − β and � EAB = α + β. Becausetan � EAC, tan � EAD, and tan � EAB form a geometric progression, it follows that

tan2 α = tan2 � EAD = tan � EAC tan � EAB = tan(α − β) tan(α + β).

By the addition formulas, we obtain

tan2 α = tan α + tan β

1 − tan α tan β· tan α − tan β

1 + tan α tan β= tan2 α − tan2 β

1 − tan2 α tan2 β,

ortan2 α − tan4 α tan2 β = tan2 α − tan2 β.

Hence, tan4 α tan2 β = tan2 β, and so tan α = 1, or α = 45◦. (We used the factthat both tan α and tan β are positive, because 0◦ < α, β < 90◦.) Thus ADE

is an isosceles right triangle with |AD| = |DE| = |AE|√2

= 5√

2. In the right

triangle ACD, |DC| = |AD| tan β, and so [ABC] = |AD| · |CD| = |AD|2 tan β =50 tan β.

Because cot � DAE = cot 45◦ = 1, cot � CAE, and cot � DAB form an arithmeticprogression, it follows that

2 cot(45◦ − β) = 2 cot � CAE = cot � DAE + cot � DAB = 1 + cot β.

Setting 45◦ − β = γ (Greek “gamma") in the above equation gives 2 cot γ =1 + cot β. Because 0◦ < β, γ < 45◦, applying the addition formulas gives

1 = cot 45◦ = cot(β + γ ) = cot β cot γ − 1

cot β + cot γ,

or cot β + cot γ = cot β cot γ − 1. Solving the system of equations{2 cot γ = 1 + cot β,

cot β + cot γ = cot β cot γ − 1or

{2 cot γ = cot β + 1,

cot γ (cot β − 1) = cot β + 1


for cot β gives (cot β+1)(cot β−1) = 2(cot β+1). It follows that cot2 β−2 cot β−3 = 0. Factoring the last equation as (cot β − 3)(cot β + 1) = 0 gives cot β = 3.Thus [ABC] = 50 tan β = 50

3 .Of course, the above solution can be simplified by using the subtraction formulas,

which will soon be developed.

Think Along the Unit Circle

Let ω denote the unit circle, that is, the circle of radius 1 centered at the originO = (0, 0). Let A be a point on ω in the first quadrant, and let θ denote the acuteangle formed by line OA and the x axis (Figure 1.10). Let A1 be the foot of theperpendicular line segment from A to the x axis. Then in the right triangle AA1O,|OA| = 1, |AA1| = sin θ , and |OA1| = cos θ . Hence A = (cos θ, sin θ).

A

A

Ox

y

x

y

��

A1 O

Figure 1.10.

In the coordinate plane, we define a standard angle (or polar angle) formed by aray � from the origin and the positive x axis as an angle through which the positivex axis can be rotated to coincide with ray �. Note that we have written a standardangle and not the standard angle. That is because there are many ways in which thepositive x axis can be rotated in order to coincide with the ray �. In particular, astandard angle of θ1 = x◦ is equivalent to a standard angle of θ2 = x◦ + k · 360◦,for all integers k. For example, a standard angle of 180◦ is equivalent to all of thesestandard angles: . . . ,−900◦, −540◦, −180◦, +540◦, +900◦, . . . . Thus a standardangle is a directed angle. By convention, a positive angle indicates rotation of thex axis in the counterclockwise direction, while a negative standard angle indicatesthat the x axis is turned in the clockwise direction.

We can also define the standard angle formed by two lines in the plane as thesmallest angle needed to rotate one line in the counterclockwise direction to coincidewith the other line. Note that this angle is always greater than or equal to 0◦ and lessthan 180◦.


For a point A in the plane, we can also describe the position of A (relative to theorigin) by the distance r = |OA| and standard angle θ formed by the line OA andthe x axis. These coordinates are called polar coordinates, and they are written inthe form A = (r, θ). (Note that the polar coordinates of a point are not unique.)

In general, for any angle θ , we define the values of sin θ and cos θ as the coordinatesof points on the unit circle. Indeed, for any θ , there is a unique point A = (x0, y0)

(in rectangular coordinates) on the unit circle ω such that A = (1, θ) (in polarcoordinates). We define cos θ = x0 and sin θ = y0; that is, A = (cos θ, sin θ) if andonly if A = (1, θ) in polar coordinates.

From the definition of the sine and cosine functions, it is clear that for all integersk, sin(θ + k · 360◦) = sin θ and cos(θ + k · 360◦) = cos θ ; that is, they are periodicfunctions with period 360◦. For θ �= (2k + 1) · 90◦, we define tan θ = sin θ

cos θ; and

for θ �= k · 180◦, we define cot θ = cos θsin θ

. It is not difficult to see that tan θ is equalto the slope of a line that forms a standard angle of θ with the x axis.

A

AB

C1

C2

D

E

xx

yy

O O

��

�

�

Figure 1.11.

Assume that A = (cos θ, sin θ). Let B be the point on ω diametrically opposite toA. Then B = (1, θ + 180◦) = (1, θ − 180◦). Because A and B are symmetric withrespect to the origin, B = (− cos θ, − sin θ). Thus

sin(θ ± 180◦) = − sin θ, cos(θ ± 180◦) = − cos θ.

It is then easy to see that both tan θ and cot θ are functions with a period of 180◦.Similarly, by rotating point A around the origin 90◦ in the counterclockwise direction(to point C2 in Figure 1.11), in the clockwise direction (to C1), reflecting across thex axis (to D), and reflecting across the y axis (to E), we can show that

sin(θ + 90◦) = cos θ, cos(θ + 90◦) = − sin θ,

sin(θ − 90◦) = − cos θ, cos(θ − 90◦) = sin θ,

sin(−θ) = − sin θ, cos(−θ) = cos θ,

sin(180◦ − θ) = sin θ, cos(180◦ − θ) = − cos θ.


Furthermore, by either reflecting A across the line y = x or using the second and thirdformulas above, we can show that sin(90◦ − θ) = cos θ and cos(90◦ − θ) = sin θ .This is the reason behind the nomenclature of the “cosine” function: “cosine” is thecomplement of sine, because the angles 90◦ −θ and θ are complementary angles. Allthese interesting and important trigonometric identities are based on the geometricproperties of the unit circle.

Earlier, we found addition and subtraction formulas defined for angles α andβ with 0◦ < α, β < 90◦ and α + β < 90◦. Under our general definitions oftrigonometry functions, we can expand these formulas to hold for all angles. Forexample, we assume that α and β are two angles with 0◦ ≤ α, β < 90◦ and α +β >

90◦. We set α′ = 90◦ − α and β ′ = 90◦ − β. Then α′ and β ′ are angles between0◦ and 90◦ with a sum of less than 90◦. By the addition formulas we developedearlier, we have

cos(α + β) = cos[180◦ − (α′ + β ′)

] = − cos(α′ + β ′)= − cos α′ cos β ′ + sin α′ sin β ′

= − cos(90◦ − α′) cos(90◦ − β ′) + sin(90◦ − α′) sin(90◦ − β ′)= − sin α sin β + cos α cos β

= cos α cos β − sin α sin β.

Thus, the addition formula for the cosine function holds for angles α and β with0◦ ≤ α, β < 90◦ and α + β > 90◦. Similarly, we can show that all the additionformulas developed earlier hold for all angles α and β. Furthermore, we can provethe subtraction formulas

sin(α − β) = sin α cos β − cos α sin β,

cos(α − β) = cos α cos β + sin α sin β,

tan(α − β) = tan α − tan β

1 + tan α tan β.

We call these, collectively, the addition and subtraction formulas. Various formsof the double-angle and triple-angle formulas are special cases of the additionand subtraction formulas. Double-angle formulas lead to various forms of the half-angle formulas. It is also not difficult to check the product-to-sum formulas bythe addition and subtraction formulas. We leave this to the reader. For angles α and


β, by the addition and subtraction formulas, we also have

sin α + sin β = sin

(α + β

2+ α − β

2

)+ sin

(α + β

2− α − β

2

)= sin

α + β

2cos

α − β

2+ cos

α + β

2sin

α − β

2

+ sinα + β

2cos

α − β

2− cos

α + β

2sin

α − β

2

= 2 sinα + β

2cos

α − β

2,

which is one of the sum-to-product formulas. Similarly, we obtain various formsof the sum-to-product formulas and difference-to-product formulas.

Example 1.4. Let a and b be nonnegative real numbers.

(a) Prove that there is a real number x such that sin x + a cos x = b if and only ifa2 − b2 + 1 ≥ 0.

(b) If sin x + a cos x = b, express |a sin x − cos x| in terms of a and b.

Solution: To establish (a), we prove a more general result.

(a) Let m, n, and � be real numbers such that m2 + n2 �= 0. We will prove thatthere is a real number x such that

m sin x + n cos x = � (∗)

if and only if m2 + n2 ≥ �2.

Indeed, we can rewrite equation (∗) in the following form:

m√m2 + n2

sin x + n√m2 + n2

cos x = �√m2 + n2

.

Point(

m√m2+n2 , n√

m2+n2

)lies on the unit circle. There is a unique real number

α with 0 ≤ α < 2π such that

cos α = m√m2 + n2

and sin α = n√m2 + n2

.

The addition and subtraction formulas yield

sin(x + α) = cos α sin x + sin α cos x = �√m2 + n2

,

which is solvable in x if and only if −1 ≤ �√m2+n2 ≤ 1, that is, if and only if

�2 ≤ m2 + n2. Setting m = a, n = 1, and � = c gives the desired result.


(b) By the relations

a2 + 1 = (sin2 x + cos2 x)(a2 + 1)

= (sin2 x + 2a sin x cos x + a2 cos2 x)

+ (a2 sin2 x − 2a sin x cos x + cos2 x)

= (sin x + a cos x)2 + (a sin x − cos x)2,

we conclude that |a sin x − cos x| = √a2 − b2 + 1.

Graphs of Trigonometric Functions

-250 -200 -150 -100 -50 50 100 150 200

1.2

1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

AB = C

x

x y

y

z z 180

-180

A1 B1C1

Figure 1.12.

We set the units of the x axis to be degrees. The graph of y = sin x looks like awave, as shown in Figure 1.12. (This is only part of the graph. The graph extendsinfinitely in both directions along the x axis.) For example, the point A = (1, x◦)corresponds to the point A1 = (x, sin x) on the curve y = sin x. If two pointsB1 and C1 are 360 from each other in the x direction, then they have the same y

value, and they correspond to the same point B = C on the unit circle. (This is thecorrespondence of the identity sin(x◦+360◦) = sin x◦.)Also, the graph is symmetricabout line x = 90. (This corresponds to the identity sin(90◦ −x◦) = sin(90◦ +x◦).)The identity sin(−x◦) = − sin x◦ indicates that the graph y = sin x is symmetricwith respect to the origin; that is, the sine is an odd function.

A function y = f (x) is sinusoidal if it can be written in the form y = f (x) =a sin[b(x + c)] + d for real constants a, b, c, and d. In particular, because cos x◦ =sin(x◦ + 90◦), y = cos x is sinusoidal (Figure 1.13). For any integer k, the graph ofy = cos x is a (90 + 360k)-unit shift to the left (or a (270 + 360k)-unit shift to theright) of the graph of y = sin x. Because cos x◦ = cos(−x◦), the cosine is an evenfunction, and so its graph is symmetric about the y axis.


-250 -200 -150 -100 -50 50 100 150 200

1.2

1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

A

A1

B

B1

x-x

x

-x

Figure 1.13.

Example 1.5. Let f be an odd function defined on the real numbers such that forx ≥ 0, f (x) = 3 sin x + 4 cos x. Find f (x) for x < 0. (See Figure 1.14.)

-250 -200 -150 -100 -50 50 100 150 200

6

4

2

-2

-4

Figure 1.14.

Solution: Because f is odd, f (x) = −f (−x). For x < 0, −x > 0 and f (−x) =3 sin(−x)+4 cos(−x) = −3 sin x+4 cos x by definition. Hence, for x < 0, f (x) =−(−3 sin x + 4 cos x) = 3 sin x − 4 cos x. (It seems that y = 3 sin x + 4 cos x mightbe sinusoidal; can you prove or disprove this?)

For a sinusoidal function y = a sin[b(x + c)] + d , it is important to note theroles played by the constants a, b, c and d in its graph. Generally speaking, a is theamplitude of the curve, b is related to the period of the curve, c is related to thehorizontal shift of the curve, and d is related to the vertical shift of the curve. To get aclearer picture, the reader might want to match the functions y = sin 3x, y = 2 cos x

3 ,y = 3 sin 4x, y = 4 cos(x − 30◦), y = 3

2 sin x2 − 3, and y = 2 sin[3(x + 40◦)] + 5

with the curves shown in Figure 1.15.


-500 -400 -300 -200 -100 100 200 300 400

15

10

5

-5

-10

-15

Figure 1.15.

We leave it to the reader to show that if a, b, c, and d are real constants, then thefunctions y = a cos(bx+c)+d, y = a sin x+b cos x, y = a sin2 x, and y = b cos2 x

are sinusoidal. Let f (x) and g(x) be two functions. For real constants a and b, thefunction af (x)+bg(x) is called a linear combination of f (x) and g(x). Is it true thatif both of f (x) and g(x) are sinusoidal, then their linear combination is sinusoidal?In fact, it is true if f and g have the same period (or frequency). We leave this proofto the reader. Figure 1.16 shows the graphs of y1 = sin x, y3 = sin x + 1

3 sin 3x,and y5 = sin x + 1

3 sin 3x + 15 sin 5x. Can you see a pattern? Back in the nineteenth

century, Fourier proved a number of interesting results, related to calculus, about thegraphs of such functions yn as n goes to infinity.

-250 -200 -150 -100 -50 50 100 150 200

2

1

-1

-2

Figure 1.16.

The graphs of y = tan x and y = cot x are not continuous, because tan x is notdefined for x = (2k + 1) · 90◦ and cot x is not defined for x = k · 180◦, where k isan integer. The graph of tan x has vertical asymptotes at x = (2k + 1) · 90◦; that is,as x approaches k · 180◦, the values of tan x grow large in absolute value, and so thegraph of the tangent function moves closer and closer to the asymptote, as shown inFigure 1.17. Similarly, the graph of cot x has vertical asymptotes at x = k · 180◦.


-400 -300 -200 -100 100 200 300

10

8

6

4

2

-2

-4

-6

-8

-10

Figure 1.17.

A function f (x) is concave up (down) on an interval [a, b] if the graph of f (x)

lies under (over) the line connecting (a1, f (a1)) and (b1, f (b1)) for all

a ≤ a1 < x < b1 ≤ b.

Functions that are concave up and down are also called convex and concave, re-spectively. In other words, the graph of a concave up function looks like a bowl thatholds water, while the graph of a concave down function looks like a bowl that spillswater.

If f is concave up on an interval [a, b] and λ1, λ2, . . ., λn (λ – Greek “lambda")are nonnegative numbers with sum equal to 1, then

λ1f (x1) + λ2f (x2) + · · · + λnf (xn) ≥ f (λ1x1 + λ2x2 + · · · + λnxn)

for any x1, x2, . . . , xn in the interval [a, b]. If the function is concave down, theinequality is reversed. This is Jensen’s inequality. Jensen’s inequality says that theoutput of a convex function at the weighted average of a group of inputs is less thanor equal to the same weighted average of the outputs of the function at the group ofinputs.

It is not difficult to see that y = sin x is concave down for 0◦ ≤ x ≤ 180◦ andy = tan x is concave up for 0◦ ≤ x < 90◦. By Jensen’s inequality, for triangle ABC,we have

1

3sin A + 1

3sin B + 1

3sin C ≤ sin

A + B + C

3=

√3

2,

or sin A + sin B + sin C ≤ 3√

32 , which is Introductory Problem 28(c). Similarly,

we have tan A + tan B + tan C ≥ 3√

3. For those who know calculus, convexityof a function is closely related to the second derivative of the function. We can alsouse the natural logarithm function to change products into sums, and then applyJensen’s inequality. This technique will certainly be helpful in solving problems


such as Introductory Problems 19(b), 20(b), 23(a) and (d), 27(b), and 28(b) and(c). Because the main goal of this book is to introduce techniques in trigonometriccomputation rather than in functional analysis, we will present solutions withoutusing Jensen’s inequality. On the other hand, we certainly do not want the reader tomiss this important method. In the second solution of Introductory Problem 51 andthe solution of Advanced Problem 39, we illustrate this technique.

The Extended Law of Sines

Let ABC be a triangle. It is not difficult to show that [ABC] = ab sin C2 . (For a proof,

see the next section.) By symmetry, we have

[ABC] = ab sin C

2= bc sin A

2= ca sin B

2.

Dividing all sides of the last equation by abc2 gives the law of sines:

sin A

a= sin B

b= sin C

cor

a

sin A= b

sin B= c

sin C.

A

B C

D

O

M

Figure 1.18.

The common ratio asin A

has a significant geometric meaning. Let ω be the cir-cumcircle of triangle ABC, and let O and R be the center and radius of ω, re-spectively. (See Figure 1.18.) Then � BOC = 2 � CAB. Let M be the midpoint ofsegment BC. Because triangle OBC is isosceles with |OB| = |OC| = R, it fol-lows that OM ⊥ BC and � BOM = � COM = � CAB. In the right triangle BMO,|BM| = |OB| sin A; that is, a

sin A= 2|BM|

sin A= 2|OB| = 2R. Hence, we obtain the

extended law of sines: In a triangle ABC with circumradius equal to R,

a

sin A= b

sin B= c

sin A= 2R.


Note that this fact can also be obtained by extending ray OB to meet ω at D, andthen working on right triangle ABD.

A direct application of the law of sines is to prove the angle-bisector theorem:Let ABC be a triangle (Figure 1.19), and let D be a point on segment BC such that� BAD = � CAD. Then

|AB||AC| = |BD|

|CD| .

Applying the law of sines to triangle ABD gives

|AB|sin � ADB

= |BD|sin � BAD

, or|AB||BD| = sin � ADB

sin � BAD.

Similarly, applying the law of sines to triangle ACD gives |AC||CD| = sin � ADC

sin � CAD. Because

sin � ADB = sin � ADC and sin � BAD = sin � CAD, it follows that |AB||BD| = |AC|

|CD| ,as desired.

This theorem can be extended to the situation in which AD1 is the external bisectorof the triangle (see Figure 1.19).We leave it to the reader to state and prove this versionof the theorem.

AA

B BC CD D1

Figure 1.19.

Area and Ptolemy’s Theorem

Let ABC be a triangle, and let D be the foot of the perpendicular line segment fromA to line BC (Figure 1.20). Then [ABC] = |BC|·|AD|

2 . Note that |AD| = |AB| sin B.

Thus [ABC] = |BC|·|AB| sin B2 = ac sin B

2 .


AA

A A B

B B

B

CC

C

C

D

DD

D

PP

P P

Figure 1.20.

In general, if P is a point on segment BC, then |AD| = |AP | sin � APB. Hence

[ABC] = |AP |·|BC| sin � APB2 . More generally, let ABCD be a quadrilateral (not

necessarily convex), and let P be the intersection of diagonals AC and BD, as shown

in Figure 1.20. Then [ABC] = |AC|·|BP | sin � APB2 and [ADC] = |AC|·|DP | sin � APD

2 .Because � APB + � APD = 180◦, it follows that sin � APB = sin � APD and

[ABCD] = [ABC] + [ADC] = |AC| sin � APB

2(|BP | + |DP |)

= |AC| · |BD| sin � APB

2.

Now we introduce Ptolemy’s theorem: In a convex cyclic quadrilateral ABCD

(that is, the vertices of the quadrilateral lie on a circle, and this circle is called thecircumcircle of the quadrilateral),

|AC| · |BD| = |AB| · |CD| + |AD| · |BC|.

There are many proofs of this very important theorem. Our proof uses areas. Theproduct |AC| · |BD| is closely related to [ABCD]. Indeed,

[ABCD] = 1

2· |AC| · |BD| sin � APB,

where P is the intersection of diagonals AC and BD. (See Figure 1.21.)


A

B

C

DP

B1

Figure 1.21.

Hence, we want to express the products |AB| · |CD| and |BC| · |DA| in terms ofareas. To do so, we reflect B across the perpendicular bisector of diagonal AC. LetB1 be the image of B under the reflection. Then ABB1C is an isosceles trapezoidwith BB1 ‖ AC, |AB| = |CB1|, and |AB1| = |CB|. Also note that B1 lies on thecircumcircle of ABCD. Furthermore, AB = CB1, and so

� B1AD = B1D

2= B1C + CD

2= AB + CD

2= � APB.

Because AB1CD is cyclic, � B1AD + � B1CD = 180◦. Thus sin � B1AD =sin � B1CD = sin � APB. Because of symmetry, we have

[ABCD] = [ABC] + [ACD] = [AB1C] + [ACD]= [AB1CD] = [AB1D] + [CB1D]= 1

2· |AB1| · |AD| sin � B1AD + 1

2· |CB1| · |CD| sin � B1CD

= 1

2· sin � APB(|BC| · |AD| + |AB| · |CD|).

By calculating [ABCD] in two different ways, we establish

1

2· |AC| · |BD| sin � APB = 1

2· sin � APB(|BC| · |AD| + |AB| · |CD|),

or |AC| · |BD| = |BC| · |AD| + |AB| · |CD|, completing the proof of the theorem.In Introductory Problem 52, we discuss many interesting properties of the special

angle 180◦7 . The following is the first of these properties.

Example 1.6. Prove that

csc180◦

7= csc

360◦

7+ csc

540◦

7.


Solution: Let α = 180◦7 . We rewrite the above equation as csc α = csc 2α + csc 3α,

orsin 2α sin 3α = sin α(sin 2α + sin 3α).

We present two approaches, from which the reader can glean both algebraic compu-tation and geometric insights.

• First Approach: Note that 3α + 4α = 180◦, so we have sin 3α = sin 4α. Itsuffices to show that

sin 2α sin 3α = sin α(sin 2α + sin 4α).

By the addition and subtraction formulas, we have sin 2α + sin 4α =2 sin 3α cos α. Then the desired result reduces to sin 2α = 2 sin α cos α, whichis the double-angle formula for the sine function.

• Second Approach: Consider a regular heptagon A1A2 . . . A7 inscribed in acircle of radius R = 1

2 (Figure 1.22). Then each arc AiAi+1 has measure360◦

7 = 2α.

A1

A2

A3

A4

A5

A6A7

Figure 1.22.

By the extended law of sines, we have |A1A2| = |A1A7| = 2R sin α = sin α,|A2A4| = |A2A7| = sin 2α, and |A1A4| = |A4A7| = sin 3α. ApplyingPtolemy’s theorem to the cyclic quadrilateral A1A2A4A7 gives

|A1A4| · |A2A7| = |A1A2| · |A4A7| + |A2A4| · |A7A1|;that is,

sin 2α sin 3α = sin α(sin 2α + sin 3α).


Existence, Uniqueness, and Trigonometric Substitutions

The fact that sin α = sin β, for α + β = 180◦, has already helped us in manyplaces. It also helped us to explain why either side-side-angle (SSA) or area-side-side information is not enough to determine the unique structure of a triangle.

Example 1.7. Let ABC be a triangle.

(a) Suppose that [ABC] = 10√

3, |AB| = 8, and |AC| = 5. Find all possiblevalues of � A.

(b) Suppose that |AB| = 5√

2, |BC| = 5√

3, and � C = 45◦. Find all possiblevalues of � A.

(c) Suppose that |AB| = 5√

2, |BC| = 5, and � C = 45◦. Find all possible valuesof � A.

(d) Suppose that |AB| = 5√

2, |BC| = 10, and � C = 45◦. Find all possiblevalues of � A.

(e) Suppose that |AB| = 5√

2, |BC| = 15, and � C = 45◦. Find all possiblevalues of � A.

Solution:

(a) Note that b = |AC| = 5, c = |AB| = 8, and [ABC] = 12bc sin A. Thus

sin A =√

32 , and A = 60◦ or 120◦ (A1 and A2 in Figure 1.23).

(b) By the law of sines, we have |BC|sin A

= |AB|sin C

, or sin A =√

32 . Hence A = 60◦ or

120◦.

(c) By the law of sines, we have |BC|sin A

= |AB|sin C

, or sin A = 12 . Hence A = 30◦ only

(A3 in Figure 1.23)! (Why?)

(d) By the law of sines, we have sin A = 1, and so A = 90◦. (A4 in Figure 1.23)

(e) By the law of sines, we have sin A = 32 , which is impossible. We conclude

that there is no triangle satisfying the conditions of the problem.


B C

A1

A2

A3

A4

Figure 1.23.

Example 1.8. [AMC12 2001] In triangle ABC, � ABC = 45◦. Point D is on segmentBC such that 2|BD| = |CD| and � DAB = 15◦. Find � ACB.

First Solution: We construct this triangle in the following way: Fix segment BC,choose point D on segment BC such that 2|BD| = |CD| (Figure 1.24, left), andconstruct ray BP such that � PBC = 45◦. Let A be a point on ray BP that movesfrom B in the direction of the ray. It is not difficult to see that � DAB decreases as A

moves away from B. Hence, there is a unique position for A such that � DAB = 15◦.This completes our construction of triangle ABC.

This figure brings to mind the proof of the angle-bisector theorem. We apply thelaw of sines to triangles ACD and ABC. Set α = � CAD. Note that � CDA =� CBA + � DAB = 60◦. We have

|CD|sin α

= |CA|sin 60◦ and

|BC|sin(α + 15◦)

= |CA|sin 45◦ .

Dividing the first equation by the second equations gives

|CD| sin(α + 15◦)|BC| sin α

= sin 45◦

sin 60◦ .

Note that |CD||BC| = 2

3 =(

sin 45◦sin 60◦

)2. It follows that

(sin 45◦

sin 60◦

)2

= sin α

sin(α + 15◦)· sin 45◦

sin 60◦ .

It is clear that α = 45◦ is a solution of the above equation. By the uniqueness of ourconstruction, it follows that � ABC = 45◦, � CAB = 60◦, and � ACB = 75◦.


A AB

C

B

C

DDE

Figure 1.24.

Second Solution: Note that � CDA = 60◦ and that sin 30◦ = 12 . We construct point

E on segment AD (Figure 1.24, right) such that CE ⊥ AD. Then in triangle CDE,� DCE = 30◦ and |DE| = |CD| sin � DCE, or |CD| = 2|DE|. Thus triangleBDE is isosceles with |DE| = |DB|, implying that � DBE = � DEB = 30◦.Consequently, � CBE = � BCE = 30◦ and � EBA = � EAB = 15◦, and sotriangles BCE and BAE are both isosceles with |CE| = |BE| = |EA|. Hencethe right triangle AEC is isosceles; that is, � ACE = � EAC = 45◦. Therefore,� ACB = � ACE + � ECB = 75◦.

For a function f : A → B, if f (A) = B, then f is said to be surjective (oronto); that is, every b ∈ B is the image under f of some a ∈ A. If every two distinctelements a1 and a2 in A have distinct images, then f is injective (or one-to-one). Iff is both injective and surjective, then f is bijective (or a bijection or a one-to-onecorrespondence).

The sine and cosine functions are functions from the set of angles to the realnumbers. The images of the two functions are the real numbers between −1 and 1.For a point P = (x, y) with polar coordinates (1, θ) on the unit circle, it is clearthat the values x = cos θ and y = sin θ vary continuously from −1 to 1, takingon all intermediate values. Hence the functions are surjective functions from theset of angles to the interval [−1, 1]. On the other hand, these two functions are notone-to-one. It is not difficult to see that the sine function is a bijection between theset of angles α with −90◦ ≤ α ≤ 90◦ and the interval [−1, 1], and that the cosinefunction is a bijection between the set of angles α with 0◦ ≤ α ≤ 180◦ and theinterval [−1, 1]. For abbreviation, we can write that sin : [−90◦, 90◦] → [−1, 1] isa bijection. It is also not difficult to see that the tangent function is a bijection betweenthe set of angles α with −90◦ < α < 90◦ (0◦ < α < 90◦, or 0◦ ≤ α < 90◦) and theset of real numbers (positive real numbers, or nonnegative real numbers).

Two functions f and g are inverses of each other if f (g(x)) = x for all x in thedomain of g and g(f (x)) = x for all x in the domain of f . If the function f is one-to-one and onto, then it is not difficult to see that f has an inverse. For a pair of functionsf and g that are inverses of each other, if y = f (x), then g(y) = g(f (x)) = x; thatis, if (a, b) lies on the graph of y = f (x), then (b, a) lies on the graph of y = g(x).


It follows that the graphs of y = f (x) and y = g(x) are reflections of each otheracross the line y = x. For real numbers x with −1 ≤ x ≤ 1, there is a unique angleθ with −90◦ ≤ θ ≤ 90◦ such that sin θ = x. Hence we define the inverse of the sinefunction, denoted by sin−1 or arcsin, in such a way that sin−1 x = θ for −1 ≤ x ≤ 1and −90◦ ≤ θ ≤ 90◦. It is important to note that sin−1 x is not (sin x)−1 or 1

sin x.

-1.5 -1 -0.5 0.5 1 1.5

100

80

60

40

20

-20

-40

-60

-80

Figure 1.25.

Similarly, we can define the inverse functions of tan x and cot x. They are denotedby tan−1 x (or arctan x) and cot−1 x. Both functions have domain R. Their rangesare {θ | −90◦ < θ < 90◦} and {θ | −90◦ < θ ≤ 90◦, θ �= 0◦}. They are bothone-to-one functions and onto functions. Their graphs are shown below. Note thaty = arctan x has two horizontal asymptotes y = 90 and y = −90. Note also thaty = cot−1 x has two pieces, and both of them are asymptotic to the line y = 0. SeeFigure 1.26.

-1.5 -1 -0.5 0.5 1 1.5

100

80

60

40

20

-20

-40

-60

-80

Figure 1.26.

Note that y = cos x is one-to-one and onto from {θ | 0◦ ≤ θ ≤ 180◦} to theinterval [−1, 1]. While the domain of cos−1 x (or arccos x) is the same as that ofarcsin x, the range of cos−1 x is {θ | 0◦ ≤ θ ≤ 180◦}. See Figure 1.27.


-1.5 -1 -0.5 0.5 1 1.5

180

160

140

120

100

80

60

40

20

Figure 1.27.

The graphs of y = sin−1 (sin x), y = cos−1 (cos x), y = sin (sin−1 x), y =cos (cos−1 x), y = cos−1 (sin x), y = sin−1 (cos x), y = sin (cos−1 x), andy = cos (sin−1 x) are interesting and important. We leave it to the reader to completethese graphs. We close this section with three examples of trigonometric substitution.

Example 1.9. Let x0 = 2003, and let xn+1 = 1+xn

1−xnfor n ≥ 1. Compute x2004.

Solution: With a little algebraic computation, we can show that this sequence hasa period of 4; that is, xn+4 = xn for all n ≥ 1. But why? We reveal the secret withtrigonometric substitution; that is, we define αn with −90◦ < αn < 90◦ such thattan αn = xn. It is clear that if xn is a real number, such an αn is unique, becausetan : (−90◦, 90◦) → R is a bijection. Because 1 = tan 45◦, we can rewrite the givencondition as

tan αn+1 = tan 45◦ + tan αn

1 − tan 45◦ tan αn

= tan(45◦ + αn),

by the addition and subtraction formulas. Consequently, αn+1 = 45◦ + αn, orαn+1 = 45◦ + αn − 180◦ (because tan has a period of 180◦). In any case, it is notdifficult to see that αn+4 = αn + k · 180◦ for some integer k. Therefore, xn+4 =tan αn+4 = tan αn = xn; that is, the sequence {xn}n≥0 has period 4, implying thatx2004 = x0 = 2003.

Example 1.10. Prove that among any five distinct real numbers there are two, a andb, such that |ab + 1| > |a − b|.

Solution: Write the numbers as tan xk , where −90◦ < xk < 90◦, k = 1, 2, 3, 4, 5.We consider the intervals (−90◦, −45◦], (−45◦, 0◦], (0◦, 45◦], and (45◦, 90◦]. Bythe pigeonhole principle, at least two of x1, x2, x3, x4, x5 lie in the same interval,say xi and xj . Then |xi − xj | < 45◦, and setting a = tan xi and b = tan xj , we get∣∣∣∣ a − b

1 + ab

∣∣∣∣ = ∣∣∣∣ tan xi − tan xj

1 + tan xi tan xj

∣∣∣∣ = | tan(xi − xj )| < tan 45◦ = 1,

and hence the conclusion follows.


Example 1.11. Let x, y, z be positive real numbers such that x+y+z = 1. Determinethe minimum value of

1

x+ 4

y+ 9

z.

Solution: An application of Cauchy–Schwarz inequality makes this is a one-stepproblem. Nevertheless, we present a proof which involves only the easier inequalityx2 + y2 ≥ 2xy for real numbers x and y, by setting first x = tan b and y = 2 tan b

and second x = tan a and y = cot a.Clearly, z is a real number in the interval [0, 1]. Hence there is an angle a such that

z = sin2 a. Then x + y = 1 − sin2 a = cos2 a, or xcos2 a

+ y

cos2 a= 1. For an angle b,

we have cos2 b + sin2 b = 1. Hence, we can set x = cos2 a cos2 b, y = cos2 a sin2 b

for some angle b. It suffices to find the minimum value of

P = sec2 a sec2 b + 4 sec2 a csc2 b + 9 csc2 a,

or

P = (tan2 a + 1)(tan2 b + 1) + 4(tan2 a + 1)(cot2 b + 1) + 9(cot2 a + 1).

Expanding the right-hand side gives

P = 14 + 5 tan2 a + 9 cot2 a + (tan2 b + 4 cot2 b)(1 + tan2 a)

≥ 14 + 5 tan2 a + 9 cot2 a + 2 tan b · 2 cot b(

1 + tan2 a)

= 18 + 9(tan2 a + cot2 a) ≥ 18 + 9 · 2 tan a cot a = 36.

Equality holds when tan a = cot a and tan b = 2 cot b, which implies that cos2 a =sin2 a and 2 cos2 b = sin2 b. Because sin2 θ + cos2 θ = 1, equality holds whencos2 a = 1

2 and cos2 b = 13 ; that is, x = 1

6 , y = 13 , z = 1

2 .

Ceva’s Theorem

A cevian of a triangle is any segment joining a vertex to a point on the opposite side.

[Ceva’s Theorem] Let AD, BE, CF be three cevians of triangle ABC. The follow-ing are equivalent (see Figure 1.28):

(1) AD, BE, CF are concurrent; that is, these lines pass a common point;

(2)sin � ABE

sin � DAB· sin � BCF

sin � EBC· sin � CAD

sin � FCA= 1;


(3)|AF ||FB| · |BD|

|DC| · |CE||EA| = 1.

We will show that (1) implies (2), (2) implies (3), and then (3) implies (1).

A

B CD

EF

P

Figure 1.28.

Assume that part (1) is true. We assume that segments AD, BE, and CF meet atpoint P . Applying the law of sines to triangle ABP yields

sin � ABE

sin � DAB= sin � ABP

sin � PAB= |AP |

|BP | .

Similarly, applying law of sines to triangle BCP and CAP gives

sin � BCF

sin � EBC= |BP |

|CP | andsin � CAD

sin � FCA= |CP |

|AP | .

Multiplying the last three identities gives part (2).Assume that part (2) is true. Applying the law of sines to triangles ABD and ACD

gives|AB||BD| = sin � ADB

sin � DABand

|DC||CA| = sin � CAD

sin � ADC.

Because � ADC + � ADB = 180◦, we have sin � ADB = sin � ADC. Multiplyingthe above identities gives

|DC||BD| · |AB|

|CA| = sin � CAD

sin � DAB.

Likewise, we have

|AE||EC| · |BC|

|AB| = sin � ABE

sin � EBCand

|BF ||FA| · |CA|

|BC| = sin � BCF

sin � FCA.

Multiplying the last three identities gives part (3).


A

B C

EF

P

D1

Figure 1.29.

Assume that part (3) is true. Let segments BE and CF meet at P , and let ray AP

meet segment BC at D1 (Figure 1.29). It suffices to show that D = D1. CeviansAD1, BE, and CF are concurrent at P . By our discussions above, we have

|AF ||FB| · |BD1|

|D1C| · |CE||EA| = 1 = |AF |

|FB| · |BD||DC| · |CE|

|EA| ,

implying that |BD1||D1C| = |BD||DC| . Because both D and D1 lie on segment BC, we conclude

that D = D1, establishing part (1).Using Ceva’s theorem, we can see that the medians, altitudes, and angle bisectors

of a triangle are concurrent. The names of these concurrent points are the centroid(G), orthocenter (H ), and incenter (I ), respectively (Figure 1.30). If the incircleof triangle ABC touches sides AB, BC, and CA at F , D, and E, then by equaltangents, we have |AE| = |AF |, |BD| = |BF |, and |CD| = |CE|. By Ceva’stheorem, it follows that lines AD, BE, and CF are concurrent, and the point ofconcurrency is called the Gergonne point (Ge) of the triangle. All these four pointsare shown in Figure 1.30. Given an angle, it is not difficult to see that the pointslying on the bisector of the angle are equidistant from the rays forming the angle.Thus, the intersection of the three angle bisectors is equidistant from the three sides.Hence, this intersection point is the center of the unique circle that is inscribed inthe triangle. That is why this point is the incenter of the triangle.

A

CB

A

B CD

E

F

I GH

Ge

Figure 1.30.

Note that Ceva’s theorem can be generalized in a such a way that the point ofconcurrency does not necessarily have to be inside the triangle; that is, the cevian


can be considered as a segment joining a vertex and a point lying on the line of theopposite side. The reader might want to establish the theorem for the configurationshown in Figure 1.31.

A

B C D

E P

F

Figure 1.31.

With this general form in mind, it is straightforward to see that in a triangle, thetwo exterior angle bisectors at two of its vertices and the interior angle bisector atthe third vertex are concurrent, and the point of concurrency is the excenter of thetriangle opposite the third vertex. Figure 1.32 shows the excenter IA of triangle ABC

opposite A. Following the reasoning of the definition of the incenter, it is not difficultto see that IA is the center of the unique circle outside of triangle ABC that is tangentto rays AB and AC and side BC.

AB

C

IA

Figure 1.32.

The following example is another good application of Ceva’s theorem.

Example 1.12. [IMO 2001 Short List] Let A1 be the center of the square inscribed inacute triangle ABC with two vertices of the square on side BC (Figure 1.33). Thusone of the two remaining vertices of the square lies on side AB and the other onsegment AC. Points B1 and C1 are defined in a similar way for inscribed squares withtwo vertices on sides AC and AB, respectively. Prove that lines AA1, BB1, CC1are concurrent.


Solution:

A

B CD E

S T

A1

A2

Figure 1.33.

Let line AA1 and segment BC intersect at A2. We define B2 and C2 analogously.By Ceva’s theorem, it suffices to show that

sin � BAA2

sin � A2AC· sin � CBB2

sin � B2BA· sin � ACC2

sin � C2CB= 1.

Let the vertices of the square be DET S, labled as shown in Figure 1.33. Applyingthe law of sines to triangles ASA1 and AT A1 gives

|AA1||SA1| = sin � ASA1

sin � SAA1= sin � ASA1

sin � BAA2and

|T A1||AA1| = sin � A1AT

sin � AT A1= sin � A2AC

sin � AT A1.

Because |A1S| = |A1T | and � ASA1 = B+45◦ and � AT A1 = C+45◦, multiplyingthe above identities yields

1 = |AA1||SA1| · |T A1|

|AA1| = sin � ASA1

sin � BAA2· sin � A2AC

sin � AT A1,

implying thatsin � BAA2

sin � A2AC= sin � ASA1

sin � AT A1= sin(B + 45◦)

sin(C + 45◦).

In exactly the same way, we can show that

sin � CBB2

sin � B2BA= sin(C + 45◦)

sin(A + 45◦)and

sin � ACC2

sin � C2CB= sin(A + 45◦)

sin(B + 45◦).

Multiplying the last three identities establishes the desired result.

Naturally, we can ask the following question: Given a triangle ABC, how doesone construct, using only a compass and straightedge, a square DEE1D1 inscribedin triangle ABC with D and E on side BC?


Think Outside the Box

A homothety (or central similarity, or dilation) is a transformation that fixes onepoint O (called its center) and maps each point P to a point P ′ for which O, P , andP ′ are collinear and the ratio OP : OP ′ = k is constant (k can be either positive ornegative). The constant k is called the magnitude of the homothety. The point P ′ iscalled the image of P , and P the preimage of P ′.

We can now answer our previous question. As shown in Figure 1.34, we first con-struct a square BCE2D2 outside of triangle ABC. (With compass and straightedge,it is possible to construct a line perpendicular to a given line. How?) Let lines AD2and AE2 meet segment BC at D and E, respectively. Then we claim that D andE are two of the vertices of the square that we are looking for. Why? If line D2E2intersects lines AB and AC at B2 and C2, then triangles ABC and AB2C2 are ho-mothetic (with center A); that is, there is a dilation centered at A that takes triangleABC, point by point, to triangle AB2C2. It is not difficult to see that the magnitudeof the homothety is |AB2||AB| = |AC2||AC| = |B2C2||BC| . Note that square BCE2D2 is inscribedin triangle AB2C2. Hence D and E, the preimages of D2 and E2, are the two desiredvertices of the inscribed square of triangle ABC.

A

B CD E

D2 E2B2 C2

S T

Figure 1.34.

Menelaus’s Theorem

While Ceva’s theorem concerns the concurrency of lines, Menelaus’s theorem isabout the collinearity of points.


[Menelaus’s Theorem] Given a triangle ABC, let F, G, H be points on lines BC,

CA, AB, respectively (Figure 1.35). Then F, G, H are collinear if and only if

|AH ||HB| · |BF |

|FC| · |CG||GA| = 1.

This is yet another application of the law of sines. Applying the law of sines totriangles AGH , BFH , and CFG yields

|AH ||GA| = sin � AGH

sin � GHA,

|BF ||HB| = sin � BHF

sin � HFB,

|CG||FC| = sin � GFC

sin � CGF.

Multiplying the last three identities gives the desired result. (Note that sin � AGH =sin � CGF , sin � BHF = sin � GHA, and sin � GFC =sin � HFB.)

A

A

B BCC F

H

G

F

H

G

Figure 1.35.

Menelaus’s theorem is very useful in geometric computations and proofs. Butmost such examples relate more to synthetic geometry approaches than trigonometriccalculations. We do not discuss such examples in this book.

The Law of Cosines

[The Law of Cosines] In a triangle ABC,

|CA|2 = |AB|2 + |BC|2 − 2|AB| · |BC| cos � ABC,

or, using standard notation,

b2 = c2 + a2 − 2ca cos B,

and analogous equations hold for |AB|2 and |BC|2.


AB

C

DA

B

C

D

Figure 1.36.

Indeed, let D be the foot of the perpendicular line segment from C to the line AB

(Figure 1.36). Then in the right triangle BCD, |BD| = a cos B and |CD| = a sin B.Hence, |DA| = |c−a cos B|; here we consider the two configurations 0◦ < A ≤ 90◦and 90◦ < A < 180◦. Then in the right triangle ACD, we have

b2 = |CA|2 = |CD|2 + |AD|2 = a2 sin2 B + (c − a cos B)2

= a2 sin2 B + c2 + a2 cos2 B − 2ac cos B

= c2 + a2 − 2ca cos B,

by noting that sin2 B + cos2 B = 1.From the length of side AB, (including) angle � ABC, and the length of side BC,

by the law of cosines, we can compute the length of the third side BC. This is calledthe SAS (side–angle–side) form of the law of cosines. On the other hand, solvingfor cos � ABC gives

cos � ABC = |AB|2 + |BC|2 − |CA|22|AB| · |BC| ,

or

cos B = c2 + a2 − b2

2ca,

and analogous formulas hold for cos C and cos A. This is the SSS (side–side–side)form of the law of cosines.

The Law of Cosines in Action, Take I: Stewart’s Theorem

[Stewart’s Theorem] Let ABC be a triangle, and let D be a point on BC (Figure1.37). Then

|BC|(|AD|2 + |BD| · |CD|

)= |AB|2 · |CD| + |AC|2 · |BD|.


A

B CDFigure 1.37.

We apply the law of cosines to triangles ABD and ACD to obtain

cos � ADB = |AD|2 + |BD|2 − |AB|22|AD| · |BD|

and

cos � ADC = |AD|2 + |CD|2 − |AC|22|AD| · |CD| .

Because � ADB + � ADC = 180◦, cos � ADB + cos � ADC = 0; that is,

|AD|2 + |BD|2 − |AB|22|AD| · |BD| + |AD|2 + |CD|2 − |AC|2

2|AD| · |CD| = 0.

Multiplying 2AD · BD · CD on both sides of the last equation gives

|CD|(|AD|2 + |BD|2 − |AB|2

)+ |BD|

(|AD|2 + |CD|2 − |AC|2

)= 0,

or|AB|2 · |CD| + |AC|2 · |BD|

= |CD|(|AD|2 + |BD|2

)+ |BD|

(|AD|2 + |CD|2

)= (|CD| + |BD|)|AD|2 + |BD| · |CD|(|BD| + |CD|)= |BC|(|AD|2 + |BD| · |CD|).

Setting D = M , the midpoint of segment BC, we can compute the length of themedian AM as a special case of Stewart’s theorem. We have

|AB|2 · |CM| + |AC|2 · |BM| = |BM|(|AM|2 + |BM| · |CM|),or

c2 · a

2+ b2 · a

2= a(|AM|2 + a

2· a

2

).

It follows that 2c2 + 2b2 = 4|AM|2 + a2, or

|AM|2 = 2b2 + 2c2 − a2

4,

which is the median formula.


The Law of Cosines in Action, Take II: Heron’s Formulaand Brahmagupta’s Formula

[Brahmagupta’s Formula] Let ABCD be a convex cyclic quadrilateral (Figure1.38). Let |AB| = a, |BC| = b, |CD| = c, |DA| = d, and s = (a + b + c + d)/2.Then

[ABCD] = √(s − a)(s − b)(s − c)(s − d).

A

B

C

Da

b

c

d

Figure 1.38.

Let B = � ABC and D = � ADC. Applying the law of cosines to triangles ABC

and DBC yields

a2 + b2 − 2ab cos B = AC2 = c2 + d2 − 2cd cos D.

Because ABCD is cyclic, B + D = 180◦, and so cos B = − cos D. Hence

cos B = a2 + b2 − c2 − d2

2(ab + cd).

It follows that

sin2 B = 1 − cos2 B = (1 + cos B)(1 − cos B)

=(

1 + a2 + b2 − c2 − d2

2(ab + cd)

)(1 − a2 + b2 − c2 − d2

2(ab + cd)

)= a2 + b2 + 2ab − (c2 + d2 − 2cd)

2(ab + cd)· c2 + d2 + 2cd − (a2 + b2 − 2ab)

2(ab + cd)

= [(a + b)2 − (c − d)2][(c + d)2 − (a − b)2]4(ab + cd)2 .

Note that

(a + b)2 − (c − d)2 = (a + b + c − d)(a + b + d − c) = 4(s − d)(s − c).


Likewise, (c + d)2 − (a − b)2 = 4(s − a)(s − b). Therefore, by observing thatB + D = 180◦ and 0◦ < B, D < 180◦, we obtain

sin B = sin D = 2√

(s − a)(s − b)(s − c)(s − d)

ab + cd,

or1

2· (ab + cd) sin B = √(s − a)(s − b)(s − c)(s − d).

Now,

[ABC] = 1

2|AB| · |BC| sin B = 1

2· ab sin B.

Likewise, we have [DBC] = 12 · cd sin B. Thus,

[ABCD] = [ABC] + [DBC] = 1

2· (ab + cd) sin B

= √(s − a)(s − b)(s − c)(s − d).

This completes the proof Brahmagupta’s formula.Further assume that there is a also circle inscribed in ABCD (Figure 1.39). Then

by equal tangents, we have a + c = b + d = s, and so [ABCD] = √abcd .

A

A

BB

C

D

C = D

Figure 1.39.

[Heron’s Formula] The area of a triangle ABC with sides a, b, c is equal to

[ABC] = √s(s − a)(s − b)(s − c),

where s = (a + b + c)/2 is the semiperimeter of the triangle.Heron’s formula can be viewed as a degenerate version of Brahmagupta’s for-

mula. Because a triangle is always cyclic, we can view triangle ABC as a cyclicquadrilateral ABCD with C = D (Figure 1.39, right); that is, CD = 0. In thisway, Brahmagupta’s formula becomes Heron’s formula. For the interested reader, itis a good exercise to prove Heron’s formula independently, following the proof ofBrahmagupta’s formula.


The Law of Cosines in Action, Take III: Brocard Points

We will show that inside any triangle ABC, there exists a unique point P (Figure1.40) such that

� PAB = � PBC = � PCA.

AB

C

P

S

Figure 1.40.

This point is called one of the two Brocard points of triangle ABC; the othersatisfies similar relations with the vertices in reverse order. Indeed, if � PAB =� PCA, then the circumcircle of triangle ACP is tangent to the line AB at A. If S isthe center of this circle, then S lies on the perpendicular bisector of segments AC,and the line SA is perpendicular to the line AB. Hence, this center can be constructedeasily. Therefore, point P lies on the circle centered at S with radius |SA| (note thatthis circle is not tangent to line BC unless |BA| = |BC|). We can use the equation� PBC = � PCA to construct the circle passing through B and tangent to line AC

at C. The Brocard point P must lie on both circles and be different from C. Such apoint is unique. The third equation � PAB = � PBC clearly holds.

AB

C

DE

F

P

Q

Figure 1.41.

We can construct the other Brocard point in a similar fashion, but in reverse order.We can also reflect lines AP , BP , and CP across the angle bisectors of � CAB,� ABC, and � BCA, respectively (Figure 1.41). Then by Ceva’s theorem, these three


new lines are also concurrent, and the point of concurrency is the second Brocardpoint. This is the reason we say that the two Brocard points are isogonal conjugatesof each other.

AB

C

P

Figure 1.42.

Example 1.13. [AIME 1999] Point P is located inside triangle ABC (Figure 1.42)so that angles PAB, PBC, and PCA are all congruent. The sides of the trianglehave lengths |AB| = 13, |BC| = 14, and |CA| = 15, and the tangent of angle PAB

is m/n, where m and n are relatively prime positive integers. Find m + n.

Solution: Let α = � PAB = � PBC = � PCA and let x, y, and z denote |PA|,|PB|, and |PC|. Apply the law of cosines to triangles PCA, PAB, and PBC toobtain

x2 = z2 + b2 − 2bz cos α,

y2 = x2 + c2 − 2cx cos α,

z2 = y2 + a2 − 2ay cos α.

Sum these three equations to obtain 2(cx + ay + bz) cos α = a2 + b2 + c2. Becausethe combined area of triangles PAB, PBC, and PCA is equal to (cx+ay+bz) sin α

2 ,the preceding equation can be rewritten as

tan α = 4[ABC]a2 + b2 + c2 .

With a = 14, b = 15, and c = 13, use Heron’s formula to find that [ABC] = 84.It follows that tan α = 168

295 , so m + n = 463.


In general, because 4[ABC] = 2ab sin C = 2bc sin A = 2ca sin B, we have

cot α = 1

tan α= a2 + b2 + c2

4[ABC] = a2

2bc sin A+ b2

2ca sin B+ c2

2ab sin C

= sin2 A

2 sin B sin C sin A+ sin2 B

2 sin C sin A sin B+ sin2 C

2 sin A sin B sin C

= sin2 A + sin2 B + sin2 C

2 sin A sin B sin C,

by the law of sines.There is another symmetric identity:

csc2 α = csc2 A + csc2 B + csc2 C.

Because � PCA + � PAC = � PAB + � PAC = � CAB, it follows that � CPA =180◦ − � CAB, and so sin � CPA = sin A. Applying the law of sines to triangleCAP gives

x

sin α= b

sin � CPA, or x = b sin α

sin A.

Similarly, by working with triangles ABP and BCP , we obtain y = c sin αsin B

andz = a sin α

sin C. Consequently,

[CAP ] = 1

2zx sin � CPA = 1

2· a sin α

sin C· b sin α

sin A· sin A

= ab sin C

2· sin2 α

sin2 C= [ABC] · sin2 α

sin2 C.

Likewise, we have [ABP ] = [ABC] · sin2 α

sin2 Aand [BCP ] = [ABC] · sin2 α

sin2 B. Adding

the last three equations gives

[ABC] = [ABC](

sin2 α

sin2 C+ sin2 α

sin2 A+ sin2 α

sin2 B

),

implying that csc2 α = csc2 A + csc2 B + csc2 C.

Vectors

In the coordinate plane, let A = (x1, y1) and B = (x2, y2). We define the vector−→AB = [x2 − x1, y2 − y1], the displacement from A to B. We use a directed segmentto denote a vector. We call the starting (or the first) point (in this case, point A) the


tail of the vector, and the ending (or the second) point (B) the head. It makes senseto write the vector

−→AC as the sum of vectors

−→AB and

−→BC, because the composite

displacements from A to B and B to C add up to the displacement from A to C.For example, as shown in Figure 1.43, left, with A = (10, 45), B = (30, 5), andC = (35, 20), then

−→AB = [20, −40], −→

BC = [5, 15], and−→AC = [25, −25].

�y

�x

�

O

�

A

�

B

�

C

��

��

��

��

��

−→AB

−→BC

−→AC

�y

�x

�

O

�

A�

B �

C

�

D

�

E

��

��

��

��

��

Figure 1.43.

In general (Figure 1.43, right), if u = [a, b] and v = [m, n], then u + v =[a+m, b+n]. If we put the tail of u at the origin, then its head is at point A = (a, b).If we also put the tail of v at the origin, then its head is at point B = (m, n). Thenu + v = −→

OA + −→OB = −→

OE, and OAEB is a parallelogram. We say that vector−→OA

is a scalar multiple of−→OC if there is a constant c such that

−→OC = [ca, cb], and c is

called the scaling factor. For abbreviation, we also write [ca, cb] as c[a, b]. If thevector

−→OA is a scalar multiple of

−→OC, it is not difficult to see that O, A, and C are

collinear. If c is positive, then we say that the two vectors point in the same direction;if c is negative, we say that the vectors point in opposite directions.

We call√

a2 + b2 the length or magnitude of vector u, and we denote it by |u|.If a �= 0, we also call b

athe slope of the vector; if a = 0, we say that u is vertical.

(These terms are naturally adapted from analytic geometry.) We say that vectors areperpendicular to each other if they form a 90◦ angle when placed tail to tail. Byproperties of slopes, we can derive that u and v are perpendicular if and only ifam + bn = 0. We can also see this fact by checking that |OA|2 + |OB|2 = |AB|2;that is, |u|2 + |v|2 = |u − v|2. It follows that u and v are perpendicular if and only if(a2 + b2) + (m2 + n2) = (a − m)2 + (b − n)2, or am + bn = 0 (Figure 1.44, left).


�y

�x

�

O

�

A

�

B

��

��

�y

�x

�

O

�

A

�

A′

�

B�

B ′

�

C′

��

��

��

� ��

��

��

��

Figure 1.44.

Note that the diagonals of a rhombus bisect the interior angles (Figure 1.44, right).

Because vectors−−→OA′ = |v|u and

−−→OB ′ = |u|v have the same length, we note that

if vectors−−→OA′,

−−→OB ′, and

−−→OC′ = −−→

OA′ + −−→OB ′ = |v|u + |u|v are placed tail to tail,−−→

OC′ bisects the angle formed by vectors−−→OA′ and

−−→OB ′, which is the same as the

angle formed by vectors u and v.A vector contains two major pieces of information: its length and its direction

(slope). Hence vectors are a very powerful tool for dealing with problems in analyticgeometry. Let’s see some examples.

Example 1.14. Alex started to wander in Wonderland at 11:00 a.m. At 12:00 p.m.,Alex was at A = (5, 26); at 1:00 p.m., Alex was spotted at B = (−7, 6). If Alexmoves along a fixed direction at a constant rate, where was Alex at 12:35 p.m.? 11:45a.m.? 1:30 p.m.? At what time and at what location did Alex cross Sesame Street,the y axis?

Solution: As shown in Figure 1.45, left, let A3, A1, A4 denote Alex’s positionsat 12:35 p.m., 11:45 p.m., and 1:30 p.m., respectively. It took Alex 60 minutesto move along the vector

−→AB = [−12, −20]. Hence he was dislocated by vector−−→

AA3 = 3560

−→AB =

[−7, − 35

3

]at 12:35 p.m. Similarly,

−−→AA1 = − 15

60−→AB = [3, 5]

and−−→AA4 = 90

60−→AB = [−18, −30]. Let O = (0, 0) be the origin. Then we find that−−→

OA3 = −→OA + −−→

AA3 = [−2, 433

]and A3 = (−2, 43

3

). Likewise, A1 = (8, 31) and

A4 = (−13, −4).Let A2 = (0, b) denote the point at which Alex crosses Sesame Street. Assume

that it took t minutes after 12:00 p.m. for Alex to cross Sesame Street. Then−−→AA2 =

t60

−−→OA1 and

−−→OA2 = −→

OA+−−→AA2; that is, [0, b] = [5, 26]+ t

60 [−12, −20]. It followsthat [0, b] = [5 − t

5 , 26 − t3

]. Solving 0 = 5 − t

5 gives t = 25, which implies that


b = 26 − 253 = 53

3 . Therefore, at 12:25 p.m., Alex crossed Sesame Street at point(0, 53

3

).

AB

A

B

MP1

P2

A1

A2

A3

A4O O

Figure 1.45.

Example 1.15. Given points A = (7, 26) and B = (12, 12), find all points P suchthat |AP | = |BP | and � APB = 90◦ (Figure 1.45, right).

Solution: Note that triangle ABP is an isosceles right triangle with |AP | = |BP |.Let M be the midpoint of segment AB. Then M = ( 19

2 , 19), |MA| = |MB| = |MP |,

and MA ⊥ MB. Thus−−→MA =

[52 , −7

], and

−−→MP =

[7, 5

2

]or

−−→MP = −

[7, 5

2

]. Let

O = (0, 0) be the origin. It follows that−→OP = −−→

OM + −−→MP = [ 19

2 , 19] ± [7, 5

2

].

Consequently, P = ( 332 , 43

2

)or P =

(52 , 33

2

).

For each of the next two examples, we present two solutions. The first solutionapplies vector operations. The second solution applies trigonometric computations.

Example 1.16. [ARML 2002] Starting at the origin, a beam of light hits a mirror (inthe form of a line) at point A = (4, 8) and is reflected to point B = (8, 12). Computethe exact slope of the mirror.

Note: The key fact in this problem is the fact that the angle of incidence is equalto the angle of reflection; that is, if the mirror lies on line PQ, as shown in Figure1.46, left, then � OAQ = � PAB.

First Solution: Construct line � such that � ⊥ PQ. Then line � bisects angle � OAB.Note that |−→AB| = √(8 − 4)2 + (12 − 8)2 = 4

√2 and |−→AO| = √

42 + 82 = 4√

5.

Thus, vector√

5 · −→AB + √

2 · −→AO bisects the angle formed by

−→AO and

−→AB; that

is, this vector and line � have the same slope. Because√

5 · −→AB + √

2 · −→AO =

1. Trigonometric Fundamentals 45[4√

5 − 4√

2, 4√

5 − 8√

2], the slope of line � is 4

√5−8

√2

4√

5−4√

2=

√5−2

√2√

5−√2

, and so the

slope of line PQ is√

5 − √2

2√

2 − √5

= (√

5 − √2)(2

√2 + √

5)

(2√

2 − √5)(2

√2 + √

5)=

√10 + 1

3.

Second Solution: Let �1 and �2 denote two lines, and for i = 1 and 2, let mi and θi

(with 0◦ ≤ θi < 180◦) denote the slope and the polar angle of line �i , respectively.Without loss of generality, we assume that θ1 > θ2. If θ is the polar angle formed bythe lines, then θ = θ1 − θ2 and

tan θ = tan θ1 − tan θ2

1 + tan θ1θ2= m1 − m2

1 − m1m2

by the addition and subtraction formulas.Let m be the slope of line PQ. Because lines OA and AB have slopes 2 and 1,

respectively, by our earlier discussion we have

m − 1

1 + m= tan � PAB = tan � QAO = 2 − m

1 + 2m,

or (m − 1)(1 + 2m) = (2 − m)(1 + m). It follows that 3m2 − 2m − 1 = 0, orm = 1±√

103 . It is not difficult to see that m = 1+√

103 is the answer to the problem.

(The other value is the slope of line �, the interior bisector of � OAB.)

A

B

Q

P

A B

M

OO

Figure 1.46.

Example 1.17. [AIME 1994] The points (0, 0), (a, 11), and (b, 37) are the verticesof an equilateral triangle (Figure 1.46, right). Find ab.

In both solutions, we set O = (0, 0), A = (a, 11), and B = (b, 37).

First Solution: Let M be the midpoint of segment AB. Then M = ( a+b2 , 24

),

OM ⊥ MA, and |OM| = √3|MA|. Because

−−→AM = ( a−b

2 , −13), it follows that


−−→OM = √

3[13, a−b

2

]. Hence,

[a+b

2 , 24] = √

3[13, a−b

2

]; that is,

a + b

2= 13

√3 and

a − b

2= 8

√3.

Adding the last two equations gives a = 21√

3, and subtracting the second equationfrom the first equation gives b = 5

√3. Consequently, ab = 315.

Second Solution: Let � α denote the angle formed by ray OA and the positivedirection of the x axis, and set x = |OA| = |OB| = |AB|. Then sin α = 11

xand

cos α = ax

. Note that ray OB forms an angle whose measure is α + 60◦ from thepositive x axis. Then by the addition and subtraction formulas, we have

37

x= sin(α + 60◦) = sin α cos 60◦ + cos α sin 60◦ = 11

2x+ a

√3

2x,

b

x= cos(α + 60◦) = cos α cos 60◦ − sin α sin 60◦ = a

2x− 11

√3

2x.

Solving the first equation for a gives a = 21√

3. We then solve the second equationfor b to obtain b = 5

√3. Hence ab = 315.

The Dot Product and the Vector Form of the Law ofCosines

In this section we introduce some basic knowledge of vector operations. Let u =[a, b] and v = [m, n] be two vectors. Define their dot product u · v = am + bn. Itis easy to check that

(i) v · v = m2 + n2 = |v|2; that is, the dot product of a vector with itself isthe square of the magnitude of v, and v · v ≥ 0 with equality if and only ifv = [0, 0];

(ii) u · v = v · u;

(iii) u · (v + w) = u · v + u · w, where w is a vector;

(iv) (cu) · v = c(u · v), where c is a scalar.

If vectors u and v are placed tail to tail at the origin O, let A and B be the heads of uand v, respectively. Then

−→AB = v − u. Let θ denote the angle formed by lines OA

and OB. Then � AOB = θ . Applying the law of cosines to triangle AOB yields

|v − u|2 = AB2 = OA2 + OB2 − 2OA · OB cos θ

= |u|2 + |v|2 − 2|u||v| cos θ.


It follows that

(v − u) · (v − u) = u · u + v · v − 2|u||v| cos θ,

or v · v − 2u · v + u · u = v · v + u · u − 2|u||v| cos θ . Hence

cos θ = u · v|u||v| .

Cauchy–Schwarz Inequality

Let u = [a, b] and v = [m, n], and let θ be the angle formed by the two vectorswhen they are placed tail to tail. Because | cos θ | ≤ 1, by the previous discussions,we conclude that (u · v)2 ≤ (|u||v|)2; that is,

(am + bn)2 ≤(a2 + b2

) (m2 + n2

).

Equality holds if and only if | cos θ | = 1, that is, if the two vectors are parallel. Inany case, the equality holds if and only if u = k · v for some nonzero real constantk; that is, a

m= b

n= k.

We can generalize the definitions of vectors for higher dimensions, and definethe dot product and the length of the vectors accordingly. This results in Cauchy–Schwarz inequality: For any real numbers a1, a2, . . . , an, and b1, b2, . . . , bn,(

a21 + a2

2 + · · · + a2n

) (b2

1 + b22 + · · · + b2

n

)≥ (a1b1 + a2b2 + · · · + anbn)

2.

Equality holds if and only if ai and bi are proportional, i = 1, 2, . . . , n.Now we revisit Example 1.11. Setting n = 3, (a1, a2, a3) = (√x,

√y,

√z),

(b1, b2, b3) =(

1√x, 2√

y, 3√

z

)in Cauchy–Schwarz inequality, we have

1

x+ 4

y+ 9

z= (x + y + z)

(1

x+ 4

y+ 9

z

)≥ (1 + 2 + 3)2 = 36.

Equality holds if and only if x1 = y

2 = z3 , or (x, y, z) = ( 1

6 , 13 , 1

2

).

Radians and an Important Limit

When a point moves along the unit circle from A = (1, 0) to B = (0, −1), it hastraveled a distance of π and through an angle of 180◦. We can use the arc length as


a way of measuring an angle. We set a conversion between units: π = 180◦; thatis, 180◦ is π radians. Hence 1 radian is equal to 180

πdegrees, which is about 57.3◦.

Therefore, an angle of α = x◦ has radian measure x · π180 , and an angle of θ = y

radians has a measure of y · 180π

degrees. To be a good problem solver, the reader isencouraged to be very familiar with the radian measures of special angles such as12◦, 15◦, 30◦, 45◦, 60◦, 120◦, 135◦, 150◦, 210◦, and vice versa.

A

B

O

Figure 1.47.

Let ω be a circle, and let O and R denote its center and radius, respectively.Suppose points A and B lie on the circle (Figure 1.47). Assume that � AOB = x◦and � AOB = y (radian measure). Then xπ

180 = y. Let |AB| denote the length of arcAB. By the symmetry of the circle, we have

|AB|2πR

= x◦

360◦ , or |AB| = xπ

180· R = yR.

Hence, if � AOB is given in radian measure, then the length of arc AB is equal tothe product of this measure and the radius of the circle. Also, by the symmetry ofthe circle, the area of this sector (the region enclosed by the circle and the two radiiOA and OB) is equal to

x◦

360◦ · πR2 = yR2

2.

That is, the area of a sector is equal to half of the product of the radian measure ofits central angle and the square of the radius of the circle.

Because radian measure reveals an important relation between the size of thecentral angle of a circle and the arc length that the angle subtends, it is a better unitby which to (algebraically) quantify a geometric object. From now on in this book,for a trigonometric function f (x), we assume that x is in radian measure, unlessotherwise specified.

Let θ be a angle with 0 < θ < π2 . We claim that

sin θ < θ < tan θ. (∗)


We consider the unit circle centered at O = (0, 0), and points A = (1, 0) and B =(cos θ, sin θ) (Figure 1.48). Then � AOB = θ . Let C be the foot of the perpendicularline segment from B to segment OA. Point D lies on ray OB such that AD ⊥ AO.Then BC = sin θ and AD = tan θ . (By our earlier discussion, arc AB has length1·θ = θ . Hence it is equivalent to show that the lengths of segments BC, arc AB, andsegment AD are in increasing order. The reader might also want to find the geometricinterpretation of cot θ , sec θ , and csc θ .) It is clear that the areas of triangle OAB,sector OAB, and triangle OAD are in increasing order; that is,

|BC| · |OA|2

<12 · θ

2<

|AD| · |OA|2

,

from which the desired result follows.

A

B

C

D

O�

Figure 1.48.

As θ approaches 0, sin θ approaches 0. We say that the limiting value of sin θ is 0as θ approaches 0, and we denote this fact by limθ→0 sin θ = 0. (The reader mightwant to explain the identity limθ→0 cos θ = limθ→0 sec θ = 1.) What about the ratio

θsin θ

? Dividing by sin θ on all sides of the inequalities in (∗) gives

1 <θ

sin θ<

1

cos θ= sec θ.

Because limθ→0

sec θ = 1, it is not difficult to see that the value of θsin θ

, which is

sandwiched in between 1 and sec θ , approaches 1 as well; that is,

limθ→0

θ

sin θ= 1, or lim

θ→0

sin θ

θ= 1.

This limit is the foundation of the computations of the derivatives of trigonometricfunctions in calculus.

Note: We were rather vague about the meaning of the term approaching. Indeed,when θ approaches 0, it can be either a small positive value or a negative value with


small magnitude. These details can be easily dealt with in calculus, which is notthe focal point of this book. We introduce this important limit only to illustrate theimportance of radian measure.

Constructing Sinusoidal Curves with a Straightedge

Example 1.18. [Phillips Exeter Academy Math Materials] Jackie wraps a sheet ofpaper tightly around a wax candle whose diameter is two units, then cuts though themboth with a sharp knife, making a 45◦ angle with the candle’s axis. After unrollingthe paper and laying it flat, Jackie sees the wavy curve formed by the cut edge,and wonders whether it can be described mathematically. Show that this curve issinusoidal.

Solution: Because we can move the paper around, without loss of generality, wemay assume that the sides of the paper meet at line AP , where A is the lowest pointon the top face of the candle after the cut, as shown in Figure 1.49. To simplify ourwork a bit more, we also pretend that the bottom 1 inch of the candle (and paperwrapping around it) was also cut off by the sharp knife, as shown in Figure 1.49.

A

P

A

B

C

D

E

FO�

Q

Figure 1.49.

Let S denote the curve formed by the cut edge before the paper is unrolled,and let D be an arbitrary point on S. For a point X on S, let X1 denote the pointcorresponding to F after the paper is unrolled. (When we unroll the paper, point A

will correspond to two points A1 and A2.) Let O be the point on the base of thecandle that is diametrically opposite to A. Let Q be the point on S such that lineQO is parallel to the axis of the candle. Set the coordinate system (on the unrolledpaper) in such a way that O1 = (0, 0), Q1 = (0, 2), and A1 = (−π, 0). Then bysymmetry, A2 = (π, 0). Let C be the midpoint of segment AO, and let ω denote the


circle centered at C with radius CA; that is, ω denotes the boundary of the base ofthe candle.

Let B be the foot of the perpendicular line segment from D to the circle ω, andassume that � OCB = θ . Because circle ω has radius 1, |OB|, the length of arc OB, isθ . (This is why we use radian measure for θ .) Then B1 = (θ, 0) and D1 = (θ, y) withy = BD. Let F be the foot of the perpendicular line segment from B to segment AC.Then CF = cos θ , and AF = 1 + cos θ . Note that A, Q, C, F , and O are coplanar,and � OAQ = 45◦. Point E is on segment AQ such that EF ⊥ AO. Consequently,� AEF = � OAQ = 45◦ and � AFE = 90◦, implying that the right triangle AEF

is isosceles, with AF = EF . It is not difficult to see that BDEF is a rectangle.Hence BD = EF = AF = 1 + cos θ . We conclude that D1 = (θ, 1 + cos θ); thatis, D1 lies on the curve y = 1 + cos x.

2

1.5

1

0.5

-0.5

-3 -2 -1 1 2 3

A1 B1

Q1

D1

A2O1

Figure 1.50.

Finally, had we not cut off the bottom of the candle, the equation of the curvewould have been y = 2 + cos x.

Three Dimensional Coordinate Systems

We view Earth as a sphere, with radius 3960 miles. We will set up two kinds of 3-Dcoordinate systems to describe the positions of places on Earth.


A

B C

D

E

FG

H

x

y

z

Figure 1.51.

The first system is the 3-D rectangular coordinate system (or Cartesian system).This is a simple generalization of the regular rectangular coordinate system in theplane (or more precisely, the xy plane). We add in the third coordinate z to describethe directed distance from a point to the xy plane. Figure 1.51 shows a rectangularbox ABCDEFGH . Note that A = (0, 0, 0), and B, D, and E are on the coordinateaxes. Given G = (6, 3, 2), we have B = (6, 0, 0), C = (6, 3, 0), D = (0, 3, 0),E = (0, 0, 2), F = (6, 0, 2), and H = (0, 3, 2). It is not difficult to see that thedistance from G to the xy, yz, zx planes, x, y, z axes, and the origin are |GC| = 2,|GH | = 6, |GF | = 3, |GB| = √

13, |GD| = 2√

10, |GE| = 3√

5, and |GA| = 7.It is not difficult to visualize this coordinate system. Just place yourself in a regular

room, choose a corner on the floor (if you are good at seeing the world upside down,you might want to try a corner on the ceiling) as the origin, and assign the three edgesgoing out of the chosen corner as the three axes. In general, for a point P = (x, y, z),x denotes the directed distance from P to the yz plane, y denotes the directed distancefrom P to the zx plane, and z denotes the directed distance from P to the xy plane.It is not difficult to see that

√x2 + y2,

√y2 + z2, and

√z2 + x2 are the respective

distances from P to the z axis, x axis, and y axis. It is also not difficult to seethat the distance between two points P1 = (x1, y1, z1) and P2 = (x2, y2, z2) is√

(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2. Based on this generalization, we can talkabout vectors in 3-D space and their lengths, and the angles formed by them whenthey are placed tail to tail. (Note that we cannot talk about the standard angle anymore.) Hence we can easily generalize the definition of the dot product of three-dimensional vectors u = [a, b, c] and v = [m, n, p] as u · v = am + bn + cp, andit is routine to check that all the properties of the dot product discussed earlier hold.

Let O be the center of Earth. We set the plane containing the equator as the xy

plane (or equatorial plane), and let the North Pole lie on the positive z axis. However,


sometimes it is not convenient to use only the rectangular system to describe positionsof places on Earth, simply because Earth is a sphere.

Example 1.19. Describe the points on Earth’s surface that can be seen from a spacestation that is 100 miles above the North Pole.

Solution: Let S denote the position of the space station, and let E be a point onEarth such that line SE is tangent to Earth’s surface. There are many such points E,and all these points form a circle C lying in a plane P that is parallel to the equatorialplane. The plane P cuts Earth’s surface into two parts. The part containing the NorthPole contains exactly those points we are looking for. The best way to describe thepoints on C is to use the angle formed by ray OE and the equatorial plane. It is notdifficult to see that as E varies along C, the angle does not change.

S

O N

E

Figure 1.52.

Let N be the point on the equatorial plane that is closest to E. The central angleEON is called the latitude of E. If E is in the northern hemisphere, then this angleis positive; otherwise it is negative. Hence the range of latitudes is [−90◦, 90◦],with 90◦ (or 90◦ north) corresponding to the North Pole and −90◦ (or 90◦ south)corresponding to the South Pole. Points of constant latitude x◦ form a circle parallelto the equator. Such a circle is called the latitudinal circle of x◦.

To solve our problem, note that cos � SOE = |OE||OS| = 3960

4060 , and so � SOE ≈12.743◦, implying that � NOE = 90◦ − � SOE ≈ 77.254◦. Therefore, all thepoints with latitude greater than or equal to 77.254◦ can be seen from the spacestation.

Example 1.20. The latitude of the town of Exeter, New Hampshire, is about 43degrees north.


(a) How far from the equatorial plane is Exeter, assuming that one travels throughEarth’s interior? What if one travels on Earth’s surface?

(b) How far does Earth’s rotation carry the citizens of Exeter during a single day?

Solution: Let N be the foot of the perpendicular line segment from E to the equa-torial plane. Then � EON = 43◦. In right triangle EON , |EN | = |EO| sin 43◦ =3960 sin 43◦ ≈ 2700.714; that is, the z coordinate of Exeter is about 2700.714miles. If one travels on Earth’s surface, we extend segment ON through N to meet

the equator at M . Then |EM| = � EON360◦ · 2π · 3960 ≈ 2971.947; that is, exeter is

about 2971.947 miles away from the equator, assuming that one travels on Earth’ssurface.

x

y

z

O

N

E

M

C

Figure 1.53.

Let C denote the latitudinal circle of 43◦, and let C be the center of C. During asingle day (and night; that is, a complete day of 24 hours), Earth’s rotation carriesthe citizens of Exeter through one revolution along the circle C. (If we fix Earth,then E travels along C for a complete revolution.) Hence, the distance sought is2π · |CE| = 2π · |ON | = 7920π cos 43◦ ≈ 18197.114 miles. (What a long freeride!)

It is not difficult to see that points E and N have the same x and y coordinates,and that N lies on a circle centered at O with radius 3960 cos 43◦. Now we set upthe x and y axes. The primary meridian is the great semicircle that passes throughGreenwich, England, on its way from the North Pole to the South Pole. The x axis isset in such a way that the intersection of the primary meridian and the equator is at(3960, 0, 0). This point is in the South Atlantic Ocean, near the coast of Ghana. They axis is set such that the positive x, y, and z axes follow the right-hand rule. Whenwe turn the primary meridian around the z axis, we obtain all the semicircles with thesegment connecting the North Pole and the South Pole as their common diameter.


These semicircles are called meridians or longitude lines. We say longitude x◦ (orlongitude line of x◦) if the standard angle of the intersection of the longitude lineand the equator is equal to x◦, and all the points on this longitude line have longitudex◦. Every point on the surface of Earth is the intersection of a latitudinal circle and alongitude line. Hence we can describe the points on the surface by an ordered pair ofangles (α, β), where α and β stand for the longitude and latitude angles, respectively.If we also consider points on any sphere, we can write E = (r, α, β), where r is theradius of the sphere. These are the spherical coordinates of the point E. For thespecial case of considering points on Earth, r is equal to 3960 miles.

x

y

z

O

N

E

M

C

��

�

O

N

Figure 1.54.

Let E be a point with spherical coordinates (r, α, β) (Figure 1.54). We com-pute its rectangular coordinates. As we have done before, let N be the foot of theperpendicular line segment from E to the equator. Let E = (x, y, z) be the rectan-gular coordinates of E. Then z = |EN | = r sin β, and N = (x, y, 0). In the xy

plane, N lies on a circle with radius r cos β, and N has standard angle α. Hencex = r cos α cos β and y = r sin α cos β; that is,

E = (r cos α cos β, r sin α cos β, r sin β).

Sometimes, we also write E = r(cos α cos β, sin α cos β, sin β). This is how toconvert spherical coordinates into rectangular coordinates. It is not difficult to reversethis procedure; that is, to convert rectangular coordinates into spherical coordinates.

Traveling on Earth

It is not possible, at least in the foreseeable future, to build tunnels through Earth’sinterior to connect big cities like New York and Tokyo. Hence, when we travel fromplace to place on Earth, we need to travel on its surface. For two places on Earth,


how can we compute the length of the shortest path along the surface connectingthe two places? Let C denote this path. Intuitively, it is not difficult to see that thepoints on C all lie on a plane. This fact is not that easy to prove, however, so let’sjust accept it. Let P denote the plane that contains C. It is clear that the intersectionof P and Earth’s surface is a circle. Consequently, we conclude that C is an arc. Fortwo fixed points, we can draw many circles passing through these two points (Figure1.55). It is not difficult to see that as the radii of the circles increase, the lengths of theminor arcs connecting the points decrease. (When the radius of the circle approachesinfinity, the circle becomes a line and the minor arc connecting the points becomesthe segment connecting the points.)

Figure 1.55.

For two points A and B on the surface of Earth, the largest circle passing throughA and B on Earth is the circle that is centered at Earth’s center. Such circles arecalled great circles. For example, the equator is a great circle, and all points withfixed longitudes form great semicircles. We encourage the reader to choose pairs ofarbitrary points on a globe and draw great circles passing through the chosen points.The reader might then have a better idea why many intercontinental airplanes fly athigh latitudes.

A

B

O

Figure 1.56.


Example 1.21. Monica and Linda traveled from Shanghai, China (east 121◦, north31◦), to Albany, NewYork (west 73◦, north 42◦), to visit their friend Hilary. Estimatethe distance of Monica and Linda’s trip.

Solution: In general, let A = (3960, α1, β1) and B = (3960, α2, β2). Then

A = 3960(cos α1 cos β1, sin α1 cos β1, sin β1)

andB = 3960(cos α2 cos β2, sin α2 cos β2, sin β2).

Let θ = � AOB. Then by the vector form of the law of cosines, we have

cos θ =−→OA · −→

OB

|−→OA||−→OB|

= cos α1 cos β1 cos α2 cos β2 + sin α1 cos β1 sin α2 cos β2 + sin β1 sin β2.

For this problem, we have α1 = 121◦, β1 = 31◦, α2 = −73◦, and β2 = 42◦.Plugging these values into the above equation gives cos θ ≈ −0.273, implyingthat θ ≈ 105.870◦. Hence the distance between Shanghai and Albany is about

θ360◦ · 2π · 3960 ≈ 7317.786 miles along Earth’s surface.

Where Are You?

During a long flight, the screens in the plane cabin often show the position of theplane and the trajectory of the trip. How is it done? Or, how does a GPS (GlobalPositioning System) work? Let’s give a baby tour of this subject.

Example 1.22. Monica and Linda traveled from Shanghai, China (east 121◦, north31◦), to Albany, NewYork (west 73◦, north 42◦), to visit their friend Hilary. Monica’shometown, Billrock, is four-fifths of the way along their trip. Find the sphericalcoordinates of the town of Billrock.

A

B

CA

B

C

D

EOO

k�

(1- )k �

Figure 1.57.


Solution: We maintain the same notation as in the solution of Example 1.21. Let C

denote an arbitrary point on AB, and assume that � AOC = k � AOB = kθ , where isk is some real number with 0 ≤ k ≤ 1. (See Figure 1.57.) Then � COB = (1 − k)θ .Let D and E be points on lines OA and OB, respectively, such that AO ‖ CE andBO ‖ CD. Then CDOE is parallelogram. Set u = −→

OA, v = −→OB, and w = −→

OC.There are real numbers a and b such that |OD| = a · |OA| and |OE| = b · |OB|.Then w = au + bv. By the distributive property of the dot product, we have

w · u = au · u + bu · v and w · v = au · v + bv · v.

Note that |u| = |v| = |w| = 3960. By the vector form of the law of cosines, theseequations are equivalent to the equations

cos kθ = a + b cos θ and cos(1 − k)θ = a cos θ + b.

Solving this system of equations for a and b gives

a = cos kθ − cos(1 − k)θ cos θ

sin2 θand b = cos(1 − k)θ − cos kθ cos θ

sin2 θ.

For our problem, we have k = 45 and θ ≈ 105.870◦ (by Example 1.21). Substituting

these values into the above equations gives a ≈ 0.376 and b ≈ 1.035. It follows thatw = au +bv ≈ 3960[0.059, −0.460, 0.886], which are the rectangular coordinatesof the point C. Let (3960, γ, φ) be the spherical coordinates of C. Then sin φ ≈0.886, or φ ≈ 62.383◦, and sin γ cos φ ≈ −0.460, or γ ≈ −82.67◦; that is, Billrockhas longitude west 82.67◦ and latitude 62.383◦.

De Moivre’s Formula

Many polynomials with real coefficients do not have real solutions, e.g., x2 +1 = 0.The traditional approach is to set i to be a number with the property i2 = −1. Tosolve an equation such as x2 − 4x + 7 = 0, we proceed as follows: (x − 2)2 = −3,or x − 2 = ±√

3i, implying that x = 2 ± √3i are the roots of the equations. (One

can also skip the step of completing the square by applying the quadratic formula.)Thus, we consider numbers in the form a+bi, where a and b are real numbers. Suchnumbers are called complex numbers or imaginary numbers. The real numberscan be viewed as complex numbers by setting a = a + 0i. Strictly speaking, thenumber i is called the imaginary unit, and z = bi is called pure imaginary if b isnonzero.

Whatever these numbers may represent, it is important to be able to visualizethem. Here is how to do it: The number a + bi is matched with the point (a, b),


or the vector [a, b] that points from the origin to (a, b). Under this convention, thecoordinate plane is called the complex plane. Points (0, b) on the y axis are therebymatched with pure imaginary numbers bi, so the y axis is called the imaginary axisin the complex plane. Similarly, the x axis is called the real axis. Let O denote theorigin, and let each lowercase letter denote the complex number assigned to the pointlabeled with the corresponding uppercase letter. (For example, if z = 3 + 4i, thenZ = (3, 4).) See Figure 1.58.

The definition of the complex plane allows us to talk about operations on complexnumbers. The sum of the complex numbers z1 = a1 + b1i and z2 = a2 + b2i isz = z1 + z2 = (a1 + a2) + (b1 + b2)i, and their difference is z′ = z1 − z2 =(a1 − a2) + (b1 − b2)i. Because of the vector interpretation of complex numbers, itis not difficult to see that OZ1ZZ2 forms a parallelogram, and z and z′ are matchedwith diagonal vectors

−→OZ and

−−−→Z2Z1.

�y

�x

� i

�

O

�

Z1 = (3, 4)z1 = 3 + 4i

�

Z2 = (−5, 2)z2 = −5 + 2i

�

Z = (−2, 6)z = −2 + 6i

�

Z′ = (8, 2)z′ = 8 + 2i

�

Q = (−5.5, 0)q = −5.5 + 0i

�

P = (−2, −3)p = −2 − 3i

�

S = (2, −3.5)s = 2 − 3.5i

��

��

��

��

��

��

��

��

��

��

��

��

��

�

��

��

Figure 1.58.

We can talk about the magnitude or length of a complex number z, denotedby |z|, by considering the magnitude of the vector it corresponds to. For example,|3−4i| = 5, and in general, |a+bi| = √

a2 + b2. Hence, all the complex numbers z

with |z| = 5 form a circle centered at the origin with radius 5. We can also talk aboutthe polar form of complex numbers. If Z = (a, b) has polar coordinates Z = (r, θ),


then r = √a2 + b2 and a = r cos θ and b = r sin θ . We write z = a + bi =

r cos θ + ri sin θ = r cis θ . For example, z = −1 + √3i = 2 cis 120◦ = 2 cis 2π

3 .We have developed interesting properties of adding vectors with equal magnitudes.

Similar properties can be written in terms of complex numbers. Let z1 and z2 becomplex numbers with |z1| = |z2| = r . Set z1 = r cis θ1 and z2 = r cis θ2. Setz = z1 + z2. Then OZ1ZZ2 is a rhombus, implying that the line OZ bisects theangle Z1OZ2; that is, z = r ′ cis θ1+θ2

2 for some real number r ′. In particular, ifz1 = 1 and z2 = cis a, then z = r ′ cis a

2 , where r ′ = OZ. Therefore, tan a2 is equal

to the slope of line OZ; that is,

tana

2= sin a

1 + cos a,

which is one of half-angle formulas. Other versions of the half-angle formulas canbe obtained in a similar fashion. It is also not difficult to see that r ′ = OZ = 2 cos a

2 .

Example 1.23. [AMC12 2002] Let a and b be real numbers such that

sin a + sin b =√

2

2,

cos a + cos b =√

6

2.

Evaluate sin(a + b).

First Solution: Square both of the given relations and add the results to obtain

sin2 a + cos2 a + sin2 b + cos2 b + 2(sin a sin b + cos a cos b) = 2

4+ 6

4,

or cos(a − b) = 2(sin a sin b + cos a cos b) = 0 by the addition and subtractionformulas. Multiplying the two given relations yields

sin a cos a + sin b cos b + sin a cos b + sin b cos a =√

3

2,

or sin 2a + sin 2b + 2 sin(a + b) = √3 by the double-angle formulas and the

addition and subtraction formulas. By the sum-to-product formulas, we obtain

sin 2a + sin 2b = 2 sin(a + b) cos(a − b) = 0.

Hence sin(a + b) =√

32 .

Second Solution:Set complex numbers z1 = cos a+ i sin a = cis a and z2 = cos b+ i sin b = cis b

(Figure 1.59).


O

Z

Z1

Z2

ab

Figure 1.59.

Then by the operation rules for complex numbers and by the given conditions,we have

z = z1 + z2 = r cisa + b

2= cis a + cis b

=√

6

2+

√2i

2= √

2

(√3

2+ i

2

)= √

2 cisπ

6;

that is, r ′ = √2 and a+b

2 = π6 , and so sin(a + b) = sin π

3 =√

32 .

One of the most interesting properties of complex numbers is related to the product.It is natural to define the product of complex numbers z1 = a1+b1i and z2 = a2+b2i

asz = z1z2 = (a1 + b1i)(a2 + b2i)

= a1a2 + a2b1i + a1b2i + b1b2i2

= (a1a2 − b1b2) + (a1b2 + a2b1)i.

This certainly mimics the form of the addition and subtraction formulas. Indeed,working in the polar form z1 = r1 cis θ1 and z2 = r2 cis θ2, we have

z = z1z2 = (a1 + b1i)(a2 + b2i) = (r1 cis θ1)(r2 cis θ2)

= r1(cos θ1 + sin θ1i)r2(cos θ2 + sin θ2i)

= r1r2(cos θ1 + sin θ1i)(cos θ2 + sin θ2i)

= r1r2[cos(θ1 + θ2) + i sin(θ1 + θ2)]= r1r2 cis(θ1 + θ2),

by noting the addition and subtraction formulas cos(θ1 + θ2) = cos θ1 cos θ2 −sin θ1 sin θ2 and sin(θ1 + θ2) = sin θ1 cos θ2 + cos θ1 sin θ2. It is then not difficult tosee that

z1

z2= r1 cis θ1

r2 cis θ2= r1

r2cis(θ1 − θ2).

We have proved the angle-addition (angle-subtraction) property of complex multi-plication (division). This is why operations on complex numbers are closely related


to trigonometry. For example, if θ1 = tan−1 12 and θ2 = tan−1 1

3 , then we can showthat θ1 + θ2 = 45◦ by the addition and subtraction formulas. On the other hand, thisfact can also be shown by simple complex multiplication: (2 + i)(3 + i) = 5 + 5i.Can the reader tell why?

By the above angle-addition property of complex multiplication, we can see that

(cos θ + i sin θ)2 = (cis θ)2 = cis 2θ,

(cos θ + i sin θ)3 = (cis θ)2(cis θ) = (cis 2θ)(cis θ) = cis 3θ,

and so on. By this induction process, we can prove de Moivre’s formula: For anyangle θ and for any integer n,

(cos θ + i sin θ)n = (cis θ)n = cis nθ = cos nθ + i sin nθ.

From this formula, it is not difficult to derive the expansion formulas of sin nα

and cos nα in terms of sin α and cos α by expanding the left-hand side of the aboveidentity and matching the corresponding real and imaginary parts of both sides:

sin nα =(

n

1

)cosn−1 α sin α −

(n

3

)cosn−3 α sin3 α

+(

n

5

)cosn−5 α sin5 α − · · · ,

cos nα =(

n

0

)cosn α −

(n

2


+(

n

4

)cosn−4 α sin4 α − · · · .

2Introductory Problems

1. Let x be a real number such that sec x − tan x = 2. Evaluate sec x + tan x.

2. Let 0◦ < θ < 45◦. Arrange

t1 = (tan θ)tan θ , t2 = (tan θ)cot θ ,

t3 = (cot θ)tan θ , t4 = (cot θ)cot θ ,

in decreasing order.

3. Compute

(a) sin π12 , cos π

12 , and tan π12 ;

(b) cos4 π24 − sin4 π

24 ;

(c) cos 36◦ − cos 72◦; and

(d) sin 10◦ sin 50◦ sin 70◦.

4. Simplify the expression√sin4 x + 4 cos2 x −

√cos4 x + 4 sin2 x.


5. Prove that

1 − cot 23◦ = 2

1 − cot 22◦ .

6. Find all x in the interval(0, π

2

)such that

√3 − 1

sin x+

√3 + 1

cos x= 4

√2.

7. Region R contains all the points (x, y) such that x2 + y2 ≤ 100 and sin(x +y) ≥ 0. Find the area of region R.

8. In triangle ABC, show that

sinA

2≤ a

b + c.

9. Let I denote the interval[−π

4 , π4

]. Determine the function f defined on the

interval [−1, 1] such that f (sin 2x) = sin x + cos x and simplify f (tan2 x)

for x in the interval I .

10. Let

fk(x) = 1

k(sink x + cosk x)

for k = 1, 2, . . . . Prove that

f4(x) − f6(x) = 1

12

for all real numbers x.

11. A circle of radius 1 is randomly placed in a 15-by-36 rectangle ABCD so thatthe circle lies completely within the rectangle. Given that the probability thatthe circle will not touch diagonal AC is m

n, where m and n are relatively prime

positive integers, find m + n.

2. Introductory Problems 65

12. In triangle ABC,

3 sin A + 4 cos B = 6 and 4 sin B + 3 cos A = 1.

Find the measure of angle C.

13. Prove thattan 3a − tan 2a − tan a = tan 3a tan 2a tan a

for all a �= kπ2 , where k is in Z.

14. Let a, b, c, d be numbers in the interval [0, π ] such that

sin a + 7 sin b = 4(sin c + 2 sin d),

cos a + 7 cos b = 4(cos c + 2 cos d).

Prove that 2 cos(a − d) = 7 cos(b − c).

15. Expresssin(x − y) + sin(y − z) + sin(z − x)

as a monomial.

16. Prove that(4 cos2 9◦ − 3)(4 cos2 27◦ − 3) = tan 9◦.

17. Prove that (1 + a

sin x

)(1 + b

cos x

)≥(

1 + √2ab)2

for all real numbers a, b, x with a, b ≥ 0 and 0 < x < π2 .

18. In triangle ABC, sin A + sin B + sin C ≤ 1. Prove that

min{A + B, B + C, C + A} < 30◦.

19. Let ABC be a triangle. Prove that


(a)

tanA

2tan

B

2+ tan

B

2tan

C

2+ tan

C

2tan

A

2= 1;

(b)

tanA

2tan

B

2tan

C

2≤

√3

9.

20. Let ABC be an acute-angled triangle. Prove that

(a) tan A + tan B + tan C = tan A tan B tan C;

(b) tan A tan B tan C ≥ 3√

3.


cot A cot B + cot B cot C + cot C cot A = 1.

Conversely, prove that if x, y, z are real numbers with xy +yz+ zx = 1, thenthere exists a triangle ABC such that cot A = x, cot B = y, and cot C = z.


sin2 A

2+ sin2 B

2+ sin2 C

2+ 2 sin

A

2sin

B

2sin

C

2= 1.

Conversely, prove that if x, y, z are positive real numbers such that

x2 + y2 + z2 + 2xyz = 1,

then there is a triangle ABC such that x = sin A2 , y = sin B

2 , and z = sin C2 .


(a) sinA

2sin

B

2sin

C

2≤ 1

8;

(b) sin2 A

2+ sin2 B

2+ sin2 C

2≥ 3

4;

(c) cos2 A

2+ cos2 B

2+ cos2 C

2≤ 9

4;

(d) cosA

2cos

B

2cos

C

2≤ 3

√3

8;


(e) cscA

2+ csc

A

2+ csc

A

2≥ 6.


(a) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C;

(b) cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C;

(c) sin2 A + sin2 B + sin2 C = 2 + 2 cos A cos B cos C;

(d) cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1.

Conversely, if x, y, z are positive real numbers such that

x2 + y2 + z2 + 2xyz = 1,

show that there is an acute triangle ABC such that x = cos A, y = cos B,C = cos C.


(a) 4R = abc

[ABC] ;

(b) 2R2 sin A sin B sin C = [ABC];(c) 2R sin A sin B sin C = r(sin A + sin B + sin C);

(d) r = 4R sinA

2sin

B

2sin

C

2;

(e) a cos A + b cos B + c cos C = abc

2R2 .

26. Let s be the semiperimeter of triangle ABC. Prove that

(a) s = 4R cosA

2cos

B

2cos

C

2;

(b) s ≤ 3√

3

2R.


(a) cos A + cos B + cos C = 1 + 4 sinA

2sin

B

2sin

C

2;


(b) cos A + cos B + cos C ≤ 3

2.


(a) cos A cos B cos C ≤ 1

8;

(b) sin A sin B sin C ≤ 3√

3

8;

(c) sin A + sin B + sin C ≤ 3√

3

2.

(d) cos2 A + cos2 B + cos2 C ≥ 3

4;

(e) sin2 A + sin2 B + sin2 C ≤ 9

4;

(f) cos 2A + cos 2B + cos 2C ≥ −3

2;

(g) sin 2A + sin 2B + sin 2C ≤ 3√

3

2.

29. Prove thattan 3x

tan x= tan

(π3

− x)

tan(π

3+ x)

for all x �= kπ6 , where k is in Z.

30. Given that

(1 + tan 1◦)(1 + tan 2◦) · · · (1 + tan 45◦) = 2n,

find n.

31. LetA = (0, 0) andB = (b, 2)be points in the coordinate plane. LetABCDEF

be a convex equilateral hexagon such that � FAB = 120◦, AB ‖ DE, BC ‖EF , and CD ‖ FA, and the y coordinates of its vertices are distinct elementsof the set {0, 2, 4, 6, 8, 10}. The area of the hexagon can be written in the formm

√n, where m and n are positive integers and n is not divisible by the square

of any prime. Find m + n.


32. Show that one can use a composition of trigonometry buttons, such as sin, cos,tan, sin−1, cos−1, and tan−1, to replace the broken reciprocal button on acalculator.

33. In triangle ABC, A − B = 120◦ and R = 8r . Find cos C.

34. Prove that in a triangle ABC,

a − b

a + b= tan

A − B

2tan

C

2.

35. In triangle ABC, ab

= 2 + √3 and � C = 60◦. Find the measure of angles A

and B.

36. Let a, b, c be real numbers, all different from −1 and 1, such that a + b + c =abc. Prove that

a

1 − a2 + b

1 − b2 + c

1 − c2 = 4abc

(1 − a2)(1 − b2)(1 − c2).

37. Prove that a triangle ABC is isosceles if and only if

a cos B + b cos C + c cos A = a + b + c

2.

38. Evaluatecos a cos 2a cos 3a · · · cos 999a,

where a = 2π1999 .

39. Determine the minimum value of

sec4 α

tan2 β+ sec4 β

tan2 α

over all α, β �= kπ2 , where k is in Z.


40. Find all pairs (x, y) of real numbers with 0 < x < π2 such that

(sin x)2y

(cos x)y2/2

+ (cos x)2y

(sin x)y2/2

= sin 2x.

41. Prove that cos 1◦ is an irrational number.

42. Find the maximum value of

S = (1 − x1)(1 − y1) + (1 − x2)(1 − y2)

if x21 + x2

2 = y21 + y2

2 = c2.

43. Prove thatsin3 a

sin b+ cos3 a

cos b≥ sec(a − b)

for all 0 < a, b < π2 .

44. If sin α cos β = − 12 , what are the possible values of cos α sin β?

45. Let a, b, c be real numbers. Prove that

(ab + bc + ca − 1)2 ≤ (a2 + 1)(b2 + 1)(c2 + 1).

46. Prove that

(sin x + a cos x)(sin x + b cos x) ≤ 1 +(

a + b

2

)2

.

47. Prove that

| sin a1| + | sin a2| + · · · + | sin an| + | cos(a1 + a2 + · · · + an)| ≥ 1.


48. Find all angles α for which the three-element set

S = {sin α, sin 2α, sin 3α}is equal to the set

T = {cos α, cos 2α, cos 3α}.

49. Let {Tn(x)}∞n=0 be the sequence of polynomials such that T0(x) = 1, T1(x) =x, Ti+1 = 2xTi(x)−Ti−1(x) for all positive integers i. The polynomial Tn(x)

is called the nth Chebyshev polynomial.

(a) Prove that T2n+1(x) and T2n(x) are odd and even functions, respectively;

(b) Prove that Tn(cos θ) = cos(nθ) for all nonnegative integers n;

(c) Prove that for real numbers x in the interval [−1, 1], −1 ≤ Tn(x) ≤ 1;

(d) Prove that Tn+1(x) > Tn(x) for real numbers x with x > 1;

(e) Determine all the roots of Tn(x);

(f) Determine all the roots of Pn(x) = Tn(x) − 1.

50. Let ABC be a triangle with � BAC = 40◦ and � ABC = 60◦. Let D and E bethe points lying on the sides AC and AB, respectively, such that � CBD = 40◦and � BCE = 70◦. Segments BD and CE meet at F . Show that AF ⊥ BC.

51. Let S be an interior point of triangle ABC. Show that at least one of � SAB,� SBC, and � SCA is less than or equal to 30◦.

52. Let a = π7 .

(a) Show that sin2 3a − sin2 a = sin 2a sin 3a;

(b) Show that csc a = csc 2a + csc 4a;

(c) Evaluate cos a − cos 2a + cos 3a;

(d) Prove that cos a is a root of the equation 8x3 + 4x2 − 4x − 1 = 0;

(e) Prove that cos a is irrational;

(f) Evaluate tan a tan 2a tan 3a;

(g) Evaluate tan2 a + tan2 2a + tan2 3a;

(h) Evaluate tan2 a tan2 2a + tan2 2a tan2 3a + tan2 3a tan2 a.

(i) Evaluate cot2 a + cot2 2a + cot2 3a.

3Advanced Problems

1. Two exercises on sin k◦ sin(k + 1)◦:

(a) Find the smallest positive integer n such that

1

sin 45◦ sin 46◦ + 1

sin 47◦ sin 48◦ + · · · + 1

sin 133◦ sin 134◦ = 1

sin n◦ .

(b) Prove that

1

sin 1◦ sin 2◦ + 1

sin 2◦ sin 3◦ + · · · + 1

sin 89◦ sin 90◦ = cos 1◦

sin2 1◦ .

2. Let ABC be a triangle, and let x be a nonnegative real number. Prove that

ax cos A + bx cos B + cx cos C ≤ 1

2(ax + bx + cx).

3. Let x, y, z be positive real numbers.


(a) Prove that

x√1 + x2

+ y√1 + y2

+ z√1 + z2

≤ 3√

3

2

if x + y + z = xyz;

(b) Prove thatx

1 − x2 + y

1 − y2 + z

1 − z2 ≥ 3√

3

2

if 0 < x, y, z < 1 and xy + yz + zx = 1.

4. Let x, y, z be real numbers with x ≥ y ≥ z ≥ π12 such that x + y + z = π

2 .Find the maximum and the minimum values of the product cos x sin y cos z.

5. Let ABC be an acute-angled triangle, and for n = 1, 2, 3, let

xn = 2n−3(cosn A + cosn B + cosn C) + cos A cos B cos C.

Prove that

x1 + x2 + x3 ≥ 3

2.

6. Find the sum of all x in the interval [0, 2π ] such that

3 cot2 x + 8 cot x + 3 = 0.

7. Let ABC be an acute-angled triangle with side lengths a, b, c and area K .Prove that√

a2b2 − 4K2 +√

b2c2 − 4K2 +√

c2a2 − 4K2 = a2 + b2 + c2

2.

8. Compute the sums(n

1

)sin a +

(n

2

)sin 2a + · · · +

(n

n

)sin na

and (n

1

)cos a +

(n

2

)cos 2a + · · · +

(n

n

)cos na.

3. Advanced Problems 75

9. Find the minimum value of

| sin x + cos x + tan x + cot x + sec x + csc x|for real numbers x.

10. Two real sequences x1, x2, . . . and y1, y2, . . . are defined in the followingway:

x1 = y1 = √3, xn+1 = xn +

√1 + x2

n, yn+1 = yn

1 +√1 + y2n

,

for all n ≥ 1. Prove that 2 < xnyn < 3 for all n > 1.

11. Let a, b, c be real numbers such that

sin a + sin b + sin c ≥ 3

2.

Prove that

sin(a − π

6

)+ sin

(b − π

6

)+ sin

(c − π

6

)≥ 0.

12. Consider any four numbers in the interval[√

2−√6

2 ,√

2+√6

2

]. Prove that there

are two of them, say a and b, such that∣∣∣a√4 − b2 − b√

4 − a2∣∣∣ ≤ 2.

13. Let a and b be real numbers in the interval [0, π2 ]. Prove that

sin6 a + 3 sin2 a cos2 b + cos6 b = 1

if and only if a = b.

14. Let x, y, z be real numbers with 0 < x < y < z < π2 . Prove that

π

2+ 2 sin x cos y + 2 sin y cos z ≥ sin 2x + sin 2y + sin 2z.


15. For a triangle XYZ, let rXYZ denote its inradius. Given that the convex pen-tagon ABCDE is inscribed in a circle, prove that if rABC = rAED andrABD = rAEC , then triangles ABC and AED are congruent.

16. All the angles in triangle ABC are less then 120◦. Prove that

cos A + cos B − cos C

sin A + sin B − sin C> −

√3

3.

17. Let ABC be a triangle such that(cot

A

2

)2

+(

2 cotB

2

)2

+(

3 cotC

2

)2

=(

6s

7r

)2

,

where s and r denote its semiperimeter and its inradius, respectively. Provethat triangle ABC is similar to a triangle T whose side lengths are all positiveintegers with no common divisor and determine these integers.

18. Prove that the average of the numbers

2 sin 2◦, 4 sin 4◦, 6 sin 6◦, . . . , 180 sin 180◦

is cot 1◦.

19. Prove that in any acute triangle ABC,

cot3 A + cot3 B + cot3 C + 6 cot A cot B cot C ≥ cot A + cot B + cot C.

20. Let {an} be the sequence of real numbers defined by a1 = t and an+1 =4an(1−an) for n ≥ 1. For how many distinct values of t do we have a2004 = 0?

21. Triangle ABC has the following property: there is an interior point P suchthat � PAB = 10◦, � PBA = 20◦, � PCA = 30◦, and � PAC = 40◦. Provethat triangle ABC is isosceles.


22. Let a0 = √2 + √

3 + √6, and let an+1 = a2

n−52(an+2)

for integers n > 0. Provethat

an = cot

(2n−3π

3

)− 2

for all n.

23. Let n be an integer with n ≥ 2. Prove that

n∏k=1

tan

[π

3

(1 + 3k

3n − 1

)]=

n∏k=1

cot

[π

3

(1 − 3k

3n − 1

)].

24. Let P2(x) = x2 − 2. Find all sequences of polynomials {Pk(x)}∞k=1 suchthat Pk(x) is monic (that is, with leading coefficient 1), has degree k, andPi(Pj (x)) = Pj (Pi(x)) for all i and j .

25. In triangleABC,a ≤ b ≤ c.As a function of angleC, determine the conditionsunder which a + b − 2R − 2r is positive, negative, or zero.

26. Let ABC be a triangle. Points D, E, F are on sides BC, CA, AB, respectively,such that |DC| + |CE| = |EA| + |AF | = |FB| + |BD|. Prove that

|DE| + |EF | + |FD| ≥ 1

2(|AB| + |BC| + |CA|).

27. Let a and b be positive real numbers. Prove that

1√1 + a2

+ 1√1 + b2

≥ 2√1 + ab

if either (1) 0 < a, b ≤ 1 or (2) ab ≥ 3.

28. Let ABC be a nonobtuse triangle such that |AB| > |AC| and � B = 45◦. LetO and I denote the circumcenter and incenter of triangle ABC, respectively.Suppose that

√2|OI | = |AB| − |AC|. Determine all the possible values of

sin A.


29. Let n be a positive integer. Find the real numbers a0 and ak,�, 1 ≤ � < k ≤ n,such that

sin2 nx

sin2 x= a0 +

∑1≤�<k≤n

ak,� cos 2(k − �)x

for all real numbers x not an integer multiple of π .

30. Let S be the set of all triangles ABC for which

5

(1

|AP | + 1

|BQ| + 1

|CR|)

− 3

min{|AP |, |BQ|, |CR|} = 6

r,

where r is the inradius and P, Q, and R are the points of tangency of theincircle with sides AB, BC, and CA, respectively. Prove that all triangles inS are isosceles and similar to one another.

31. Let a, b, c be real numbers in the interval (0, π2 ). Prove that

sin a sin(a − b) sin(a − c)

sin(b + c)+ sin b sin(b − c) sin(b − a)

sin(c + a)

+ sin c sin(c − a) sin(c − b)

sin(a + b)≥ 0.


sin3A

2+ sin

3B

2+ sin

3C

2≤ cos

A − B

2+ cos

B − C

2+ cos

C − A

2.

33. Let x1, x2, . . . , xn, n ≥ 2, be n distinct real numbers in the interval [−1, 1].Prove that

1

t1+ 1

t2+ · · · + 1

tn≥ 2n−2,

where ti =∏j �=i |xj − xi |.

34. Let x1, . . . , x10 be real numbers in the interval [0, π/2] such that sin2 x1 +sin2 x2 + · · · + sin2 x10 = 1. Prove that

3(sin x1 + · · · + sin x10) ≤ cos x1 + · · · + cos x10.


35. Let x1, x2, . . . , xn be arbitrary real numbers. Prove the inequality

x1

1 + x21

+ x2

1 + x21 + x2

2

+ · · · + xn

1 + x21 + · · · + x2

n

<√

n.

36. Let a0, a1, . . . , an be numbers in the interval(0, π

2

)such that

tan(a0 − π

4

)+ tan

(a1 − π

4

)+ · · · + tan

(an − π

4

)≥ n − 1.

Prove thattan a0 tan a1 · · · tan an ≥ nn+1.

37. Find all triples of real numbers (a, b, c) such that a2−2b2 = 1, 2b2−3c2 = 1,and ab + bc + ca = 1.

38. Let n be a positive integer, and let θi be angles with 0 < θi < 90◦ such that

cos2 θ1 + cos2 θ2 + · · · + cos2 θn = 1.

Prove that

tan θ1 + tan θ2 + · · · + tan θn ≥ (n − 1)(cot θ1 + cot θ2 + · · · + cot θn).

39. One of the two inequalities

(sin x)sin x < (cos x)cos x and (sin x)sin x > (cos x)cos x

is always true for all real numbers x such that 0 < x < π4 . Identify that

inequality and prove your result.

40. Let k be a positive integer. Prove that√

k + 1 − √k is not the real part of the

complex number z with zn = 1 for some positive integer n.

41. Let A1A2A3 be an acute-angled triangle. Points B1, B2, B3 are on sides A2A3,

A3A1, A1A2, respectively. Prove that

2(b1 cos A1 + b2 cos A2 + b3 cos A3) ≥ a1 cos A1 + a2 cos A2 + a3 cos A3,

where ai = |Ai+1Ai+2| and bi = |Bi+1Bi+2|, for i = 1, 2, 3 (with indicestaken modulo 3; that is, xi+3 = xi).


42. Let ABC be a triangle. Let x, y, and z be real numbers, and let n be a positiveinteger. Prove the following four inequalities.

(a) x2 + y2 + z2 ≥ 2yz cos A + 2zx cos B + 2xy cos C.

(b) x2 + y2 + z2 ≥ 2(−1)n+1(yz cos nA + zx cos nB + xy cos nC).

(c) yza2 + zxb2 + xyc2 ≤ R2(x + y + z)2.

(d) xa2 + yb2 + zc2 ≥ 4[ABC]√xy + yz + zx.

43. A circle ω is inscribed in a quadrilateral ABCD. Let I be the center of ω.Suppose that

(|AI | + |DI |)2 + (|BI | + |CI |)2 = (|AB| + |CD|)2.

Prove that ABCD is an isosceles trapezoid.

44. Let a, b, and c be nonnegative real numbers such that

a2 + b2 + c2 + abc = 4.

Prove that0 ≤ ab + bc + ca − abc ≤ 2.

45. Let s, t, u, v be numbers in the interval(0, π

2

)with s + t + u + v = π . Prove

that√

2 sin s − 1

cos s+

√2 sin t − 1

cos t+

√2 sin u − 1

cos u+

√2 sin v − 1

cos v≥ 0.

46. Suppose a calculator is broken and the only keys that still work are the sin, cos,tan, sin−1, cos−1, and tan−1 buttons. The display initially shows 0. Given anypositive rational number q, show that we can get q to appear on the displaypanel of the calculator by pressing some finite sequence of buttons. Assumethat the calculator does real-number calculations with infinite precision, andthat all functions are in terms of radians.


47. Let n be a fixed positive integer. Determine the smallest positive real numberλ such that for any θ1, θ2, . . . , θn in the interval

(0, π

2

), if

tan θ1 tan θ2 · · · tan θn = 2n/2,

thencos θ1 + cos θ2 + · · · + cos θn ≤ λ.

48. Let ABC be an acute triangle. Prove that

(sin 2B + sin 2C)2 sin A + (sin 2C + sin 2A)2 sin B

+ (sin 2A + sin 2B)2 sin C ≤ 12 sin A sin B sin C.

49. On the sides of a nonobtuse triangle ABC are constructed externally a squareP4, a regular m-sided polygon Pm, and a regular n-sided polygon Pn. Thecenters of the square and the two polygons form an equilateral triangle. Provethat m = n = 6, and find the angles of triangle ABC.

50. Let ABC be an acute triangle. Prove that(cos A

cos B

)2

+(

cos B

cos C

)2

+(

cos C

cos A

)2

+ 8 cos A cos B cos C ≥ 4.

51. For any real number x and any positive integer n, prove that∣∣∣∣∣n∑

k=1

sin kx

k

∣∣∣∣∣ ≤ 2√

π.

4Solutions to Introductory Problems

1. [AMC12 1999] Let x be a real number such that sec x − tan x = 2. Evaluatesec x + tan x.

Solution: Note that

1 = sec2 x − tan2 x = (sec x + tan x)(sec x − tan x).

Hence sec x + tan x = 12 .

2. Let 0◦ < θ < 45◦. Arrange

t1 = (tan θ)tan θ , t2 = (tan θ)cot θ ,

t3 = (cot θ)tan θ , t4 = (cot θ)cot θ ,

in decreasing order.

Solution: For a > 1, the function y = ax is an increasing function. For0◦ < θ < 45◦, cot θ > 1 > tan θ > 0. Thus t3 < t4.

For a < 1, the function y = ax is a decreasing function. Thus t1 > t2.

Again, by cot θ > 1 > tan θ > 0, we have t1 < 1 < t3. Hence t4 > t3 > t1 >

t2.


3. Compute

(a) sin π12 , cos π

12 , and tan π12 ;

(b) cos4 π24 − sin4 π

24 ;

(c) cos 36◦ − cos 72◦; and

(d) sin 10◦ sin 50◦ sin 70◦.

Solution:

(a) By the double-angle and addition and subtraction formulas, we have

cosπ

12= cos

(π3

− π

4

)= cos

π

3cos

π

4+ sin

π

3sin

π

4

= 1

2·√

2

2+

√3

2·√

2

2=

√2 + √

6

4.

Similarly, we can show that sin π12 =

√6−√

24 . It follows that tan π

12 =√6−√

2√6+√

2= 2 − √

3.

(b) By the double-angle and addition and subtraction formulas, we obtain

cos4 π

24− sin4 π

24=(

cos2 π

24+ sin2 π

24

) (cos2 π

24− sin2 π

24

)= 1 · cos

π

12=

√2 + √

6

4.

(c) Note that

cos 36◦ − cos 72◦ = 2(cos 36◦ − cos 72◦)(cos 36◦ + cos 72◦)2(cos 36◦ + cos 72◦)

= 2 cos2 36◦ − 2 cos2 72◦

2(cos 36◦ + cos 72◦).

By the double-angle formulas, the above equality becomes

cos 36◦ − cos 72◦ = cos 72◦ + 1 − cos 144◦ − 1

2(cos 36◦ + cos 72◦)

= cos 72◦ + cos 36◦

2(cos 36◦ + cos 72◦)= 1

2.

This fact can also be seen in an isosceles triangleABC with AB = AC,BC = 1, and A = 36◦. Point D lies on side AC with � ABD = � DBC.We leave it to the reader to show that BC = BD = AD = 1, AB =2 cos 36, and CD = 2 cos 72, from which the desired result follows.

4. Solutions to Introductory Problems 85

(d) Applying the double-angle formulas again gives

8 sin 20◦ sin 10◦ sin 50◦ sin 70◦ = 8 sin 20◦ cos 20◦ cos 40◦ cos 80◦

= 4 sin 40◦ cos 40◦ cos 80◦

= 2 sin 80◦ cos 80◦

= sin 160◦ = sin 20◦.Consequently,

sin 10◦ sin 50◦ sin 70◦ = 1

8.

4. [AMC12P 2002] Simplify the expression√sin4 x + 4 cos2 x −

√cos4 x + 4 sin2 x.

Solution: The given expression is equal to√sin4 x + 4(1 − sin2 x) −

√cos4 x + 4(1 − cos2 x)

=√

(2 − sin2 x)2 −√

(2 − cos2 x)2 = (2 − sin2 x) − (2 − cos2 x)

= cos2 x − sin2 x = cos 2x.

5. Prove that

1 − cot 23◦ = 2

1 − cot 22◦ .

First Solution: We will show that

(1 − cot 23◦)(1 − cot 22◦) = 2.

Indeed, by the addition and subtraction formulas, we obtain

(1 − cot 23◦)(1 − cot 22◦) =(

1 − cos 23◦

sin 23◦

)(1 − cos 22◦

sin 22◦

)= sin 23◦ − cos 23◦

sin 23◦ · sin 22◦ − cos 22◦

sin 22◦

=√

2 sin(23◦ − 45◦)√

2 sin(22◦ − 45◦)sin 23◦ · sin 22◦

= 2 sin(−22◦) sin(−23◦)sin 23◦ sin 22◦

= 2 sin 22◦ sin 23◦

sin 23◦ sin 22◦ = 2.


Second Solution: Note that by the addition and subtraction formula, we have

cot 22◦ cot 23◦ − 1

cot 22◦ + cot 23◦ = cot(22◦ + 23◦) = cot 45◦ = 1.

Hence cot 22◦ cot 23◦ − 1 = cot 22◦ + cot 23◦, and so

1 − cot 22◦ − cot 23◦ + cot 22◦ cot 23◦ = 2,

that is, (1 − cot 23◦)(1 − cot 22◦) = 2, as desired.

6. Find all x in the interval(0, π

2

)such that

√3 − 1

sin x+

√3 + 1

cos x= 4

√2.

Solution: From Problem 3(a), we have cos π12 =

√2+√

64 and sin π

12 =√

6−√2

4 .Write the given equation as

√3−14

sin x+

√3+14

cos x= √

2,

orsin π

12

sin x+ cos π

12

cos x= 2.

Clearing the denominator gives

sinπ

12cos x + cos

π

12sin x = 2 sin x cos x,

or sin(

π12 + x

) = sin 2x. We obtain π12 + x = 2x and π

12 + x = π − 2x,implying that x = π

12 and x = 11π36 . Both solutions satisfy the given condition.

7. Region R contains all the points (x, y) such that x2 + y2 ≤ 100 and sin(x +y) ≥ 0. Find the area of region R.

Solution: Let C denote the disk (x, y) with x2 + y2 ≤ 100. Because sin(x +y) = 0 if and only if x +y = kπ for integers k, disk C has been cut by parallellines x+y = kπ , and in between those lines there are regions containing points(x, y) with either sin(x + y) > 0 or sin(x + y) < 0. Since sin(−x − y) =− sin(x + y), the regions containing points (x, y) with sin(x + y) > 0 aresymmetric with respect to the origin to the regions containing points (x, y)

with sin(x + y) < 0. Thus, as indicated in Figure 4.1, the area of region R ishalf the area of disk C, that is, 50π .


Figure 4.1.


sinA

2≤ a

b + c.

Solution: By the extended law of sines, we have

a

b + c= sin A

sin B + sin C.

Applying the double-angle formulas and sum-to-product formulas in theabove relation gives

a

b + c= 2 sin A

2 cos A2

2 sin B+C2 cos B−C

2

= sin A2

cos B−C2

≥ sinA

2,

by noting that 0 < cos B−C2 ≤ 1, because 0 ≤ |B − C| < 180◦.

Note: By symmetry, we have analogous formulas

sinB

2≤ b

c + aand sin

C

2≤ c

a + b.

9. Let I denote the interval [−π4 , π

4 ]. Determine the function f defined on theinterval [−1, 1] such that f (sin 2x) = sin x + cos x and simplify f (tan2 x)

for x in the interval I .


Solution: Note that

[f (sin 2x)]2 = (sin x + cos x)2

= sin2 x + cos2 x + 2 sin x cos x

= 1 + sin 2x.

Note also that sin 2x is a one-to-one and onto function from I to the interval[−1, 1], that is, for every −1 ≤ t ≤ 1, there is a unique x in I such that sin 2x =t . Hence, for −1 ≤ t ≤ 1, [f (t)]2 = 1 + t . For x in I , sin x + cos x ≥ 0.Therefore, f (t) = √

1 + t for −1 ≤ t ≤ 1.

For π4 ≤ x ≤ π

4 , −1 ≤ tan x ≤ 1, and so 0 ≤ tan2 x ≤ 1. Thus,

f (tan2 x) =√

1 + tan2 x = sec x.

10. Let

fk(x) = 1

k(sink x + cosk x)

for k = 1, 2, . . . . Prove that

f4(x) − f6(x) = 1

12

for all real numbers x.

Solution: We need to show that

3(sin4 x + cos4 x) − 2(sin6 x + cos6 x) = 1

for all real numbers x. Indeed, the left-hand side is equal to

3[(sin2 x + cos2 x)2 − 2 sin2 x cos2 x]− 2(sin2 x + cos2 x)(sin4 x − sin2 x cos2 x + cos4 x)

= 3 − 6 sin2 x cos2 x − 2[(sin2 x + cos2 x)2 − 3 sin2 x cos2 x]= 3 − 2 = 1.


11. [AIME 2004, by Jonathan Kane] A circle of radius 1 is randomly placed in a15×36 rectangle ABCD so that the circle lies completely within the rectangle.Compute the probability that the circle will not touch diagonal AC.

Note: In order for the circle to lie completely within the rectangle, the centerof the circle must lie in a rectangle that is (15 − 2) × (36 − 2), or 13 × 34.The requested probability is equal to the probability that the distance from thecircle’s center to the diagonal AC is greater than 1, which equals the probabilitythat the distance from a randomly selected point in the 13 × 34 rectangle toeach side of triangles ABC and CDA is greater than 1. Let |AB| = 36 and|BC| = 15 (and so |AC| = 39). Draw three segments that are 1 unit away fromeach side of triangle ABC and whose endpoints are on the sides. Let E, F, andG be the three points of intersection nearest to A, B, and C, respectively, of thethree segments. Because the corresponding sides of triangle ABC and EFG

are parallel, the two triangles are similar to each other. The desired probabilityis equal to

2[EFG]13 · 34

=( |EF |

|AB|)2

· 2[ABC]13 · 34

=( |EF |

|AB|)2

· 15 · 36

13 · 34=( |EF |

|AB|)2

· 270

221.

Because E is equidistant from sides AB and AC, E lies on the bisector of� CAB. Similarly, F and G lie on the bisectors of � ABC and � BCA, respec-tively. Hence lines AE, BF , and CG meet I , the incenter of triangle ABC.

A B

CD

E F

G

I

E1 F1

Figure 4.2.

First Solution: Let E1 and F1 be the feet of the perpendiculars from E andF to segment AB, respectively. Then |EF | = |E1F1|. It is not difficult to seethat |BF1| = |FF1| = |EE1| = 1. Set θ = � EAB. Then � CAB = 2θ ,sin 2θ = 5

12 , cos 2θ = 1213 , and tan 2θ = 5

12 . By either the double-angleformulas or the half-angle formulas,

tan 2θ = 2 tan θ

1 − tan2 θor tan θ = 1 − cos 2θ

sin 2θ,


and we obtain tan θ = 15 . It follows that |EE1||AE1| = tan θ = 1

5 , or |AE1| = 5.

Consequently, |EF | = |E1F1| = 30. Hence mn

= ( 3036

)2 · 270221 = 375

442 , andm + n = 817.

Second Solution: Set A = (0, 0), B = (36, 0), and C = (36, 15). BecauseE lies on the angle bisector of � CAB,

−→AE has the same slope as |−→AB|−→AC +

|−→AC|−→AB = 36[36, 15] + 39[36, 0] = [75 · 36, 36 · 15] = 36 · 15[5, 1]; thatis, the slope of line AE is 1

5 . Consequently, |EE1| = 5, and the rest of thesolution proceeds like that of the first solution.

Third Solution: Because the corresponding sides of triangles ABC and EFG

are parallel, it follows that I is also the incenter of triangle EFG and thatthe triangles are homothetic (with I as the center). If r is the inradius oftriangle ABC, then r − 1 is the inradius of triangle EFG; that is, the ratioof the similarity between triangles EFG and ABC is r−1

r. Hence the desired

probability is(

r−1r

)2 · 270221 .

Note that r(|AB|+|BC|+|CA|) = 2([AIB]+[BIC]+[CIA]) = 2[ABC] =|AB| · |BC|. Solving the last equation gives r = 6, and so

(56

)2 · 270221 = 375

442 .

12. [AMC12 1999] In triangle ABC,

3 sin A + 4 cos B = 6 and 4 sin B + 3 cos A = 1.

Find the measure of angle C.

Solution: Square the two given equations and add the results to obtain

24(sin A cos B + cos A sin B) = 12,

or sin(A+B) = 12 . Because C = 180◦−A−B, we have sin C = sin(A+B) =

12 , implying that either C = 30◦ or C = 150◦. But if C = 150◦, A < 30◦, so3 sin A+4 cos B < 3

2 +4 < 6, a contradiction. Hence the answer is C = 30◦.

13. Prove thattan 3a − tan 2a − tan a = tan 3a tan 2a tan a

for all a �= kπ2 , where k is in Z.


Solution: The equality is equivalent to

tan 3a(1 − tan 2a tan a) = tan 2a + tan a,

or

tan 3a = tan 2a + tan a

1 − tan 2a tan a.

That is, tan 3a = tan(2a + a), which is evident.

Note: More generally, if a1, a2, a3 are real numbers different from kπ2 , where

k is in Z, such that a1 + a2 + a3 = 0, then the relation

tan a1 + tan a2 + tan a3 = tan a1 tan a2 tan a3

holds. The proof of this relation is similar to the proofs of Problems 13 and20. We leave the proof as an exercise for the reader.

14. Let a, b, c, d be numbers in the interval [0, π ] such that

sin a + 7 sin b = 4(sin c + 2 sin d),

cos a + 7 cos b = 4(cos c + 2 cos d).

Prove that 2 cos(a − d) = 7 cos(b − c).

Solution: Rewrite the two given equalities as

sin a − 8 sin d = 4 sin c − 7 sin b,

cos a − 8 cos d = 4 cos c − 7 cos b.

By squaring the last two equalities and adding them, we obtain

1+64−16(cos a cos d +sin a sin d) = 16+49−56(cos b cos c+sin b sin c),

and the conclusion follows from the addition formulas.

15. Expresssin(x − y) + sin(y − z) + sin(z − x)

as a monomial.


Solution: By the sum-to-product formulas, we have

sin(x − y) + sin(y − z) = 2 sinx − z

2cos

x + z − 2y

2.

By the double-angle formulas, we have

sin(z − x) = 2 sinz − x

2cos

z − x

2.

Thus,sin(x − y) + sin(y − z) + sin(z − x)

= 2 sinx − z

2

[cos

x + z − 2y

2− cos

z − x

2

]= −4 sin

x − z

2sin

z − y

2sin

x − y

2

= −4 sinx − y

2sin

y − z

2sin

z − x

2by the sum-to-product formulas.

Note: In exactly the same way, we can show that if a, b, and c are real numberswith a + b + c = 0, then

sin a + sin b + sin c = −4 sina

2sin

b

2sin

c

2.

In Problem 15, we have a = x − y, b = y − z, and c = z − x.

16. Prove that(4 cos2 9◦ − 3)(4 cos2 27◦ − 3) = tan 9◦.

Solution: We have cos 3x = 4 cos3 x − 3 cos x, so 4 cos2 x − 3 = cos 3xcos x

forall x �= (2k + 1) · 90◦, k ∈ Z. Thus

(4 cos2 9◦ − 3)(4 cos2 27◦ − 3) = cos 27◦

cos 9◦ · cos 81◦

cos 27◦ = cos 81◦

cos 9◦

= sin 9◦

cos 9◦ = tan 9◦,

as desired.


17. Prove that (1 + a

sin x

)(1 + b

cos x

)≥(

1 + √2ab)2

for all real numbers a, b, x with a, b ≥ 0 and 0 < x < π2 .

Solution: Expanding both sides, the desired inequality becomes

1 + a

sin x+ b

cos x+ ab

sin x cos x> 1 + 2ab + 2

√2ab.

By the arithmetic–geometric means inequality, we obtain

a

sin x+ b

cos x≥ 2

√ab√

sin x cos x.

By the double-angle formulas, we have sin x cos x = 12 sin 2x ≤ 1

2 , and so

2√

ab√sin x cos x

≥ 2√

2ab

andab

sin x cos x≥ 2ab.

Combining the last three inequalities gives the the desired result.

18. In triangle ABC, sin A + sin B + sin C ≤ 1. Prove that

min{A + B, B + C, C + A} < 30◦.

Solution: Without loss of generality, we assume that A ≥ B ≥ C. Weneed to prove that B +C < 30◦. The law of sines and the triangle inequality(b+c > a) imply that sin B+sin C > sin A, so sin A+sin B+sin C > 2 sin A.It follows that sin A < 1

2 , and the inequality A ≥ A+B+C3 = 60◦ gives that

A > 150◦; that is, B + C < 30◦, as desired.


(a)

tanA

2tan

B

2+ tan

B

2tan

C

2+ tan

C

2tan

A

2= 1;


(b)

tanA

2tan

B

2tan

C

2≤

√3

9.

Solution: By the addition and subtraction formulas, we have

tanA

2+ tan

B

2= tan

A + B

2

(1 − tan

A

2tan

B

2

).

Because A + B + C = 180◦, A+B2 = 90◦ − C

2 , and so tan A+B2 = cot C

2 .Thus,

tanA

2tan

B

2+ tan

B

2tan

C

2+ tan

C

2tan

A

2

= tanA

2tan

B

2+ tan

C

2cot

C

2

(1 − tan

A

2tan

B

2

)= tan

A

2tan

B

2+ 1 − tan

A

2tan

B

2= 1,

establishing (a).

By the arithmetic–geometric means inequality, we have

1 = tanA

2tan

B

2+ tan

B

2tan

C

2+ tan

C

2tan

A

2

≥ 33

√(tan

A

2tan

B

2tan

C

2

)2

,

from which (b) follows.

Note: An equivalent form of (a) is

cotA

2+ cot

B

2+ cot

C

2= cot

A

2cot

B

2cot

C

2.

20. Let ABC be an acute-angled triangle. Prove that

(a) tan A + tan B + tan C = tan A tan B tan C;

(b) tan A tan B tan C ≥ 3√

3.


Solution: Note that because of the condition A, B, C �= 90◦, all the aboveexpressions are well defined.

The proof of the identity in part (a) is similar to that of Problem 19(a). By thearithmetic–geometric means inequality,

tan A + tan B + tan C ≥ 3 3√

tan A tan B tan C.

By (a), we have

tan A tan B tan C ≥ 3 3√

tan A tan B tan C,


Note: Indeed, the identity in (a) holds for all angles A, B, C with A+B+C =mπ and A, B, C �= kπ

2 , where k and m are in Z.


cot A cot B + cot B cot C + cot C cot A = 1.

Conversely, prove that if x, y, z are real numbers with xy +yz+ zx = 1, thenthere exists a triangle ABC such that cot A = x, cot B = y, and cot C = z.

Solution: If ABC is a right triangle, then without loss of generality, assumethat A = 90◦. Then cot A = 0 and B + C = 90◦, and so cot B cot C = 1,implying the desired result.

If A, B, C �= 90◦, then tan A tan B tan C is well defined. Multiplying bothsides of the desired identity by tan A tan B tan C reduces the desired result toIntroductory Problem 20(a).

The second claim is true because cot x is a bijective function from the interval(0◦, 180◦) to (−∞, ∞).


sin2 A

2+ sin2 B

2+ sin2 C

2+ 2 sin

A

2sin

B

2sin

C

2= 1.

Conversely, prove that if x, y, z are positive real numbers such that

x2 + y2 + z2 + 2xyz = 1,


then there is a triangle ABC such that x = sin A2 , y = sin B

2 , and z = sin C2 .

Solution: Solving the second given equation as a quadratic in x gives

x = −2yz +√4y2z2 − 4(y2 + z2 − 1)

2= −yz +

√(1 − y2)(1 − z2).

We make the trigonometric substitution y = sin u and z = sin v, where0◦ < u, v < 90◦. Then

x = − sin u sin v + cos u cos v = cos (u + v).

Set u = B2 , v = C

2 , and A = 180◦ − B − C. Because 1 ≥ y2 + z2 =sin2 B

2 + sin2 C2 , cos2 B

2 ≥ sin2 C2 . Because 0◦ < B

2 , C2 < 90◦, cos B

2 >

sin C2 = cos

(90◦ − C

2

), implying that B

2 < 90◦ − C2 , or B + C < 180◦. Then

x = cos(u + v) = sin A2 , y = sin B

2 , and z = sin C2 , where A, B, and C are

the angles of a triangle.

If ABC is a triangle, all the above steps can be reversed to obtain the firstgiven identity.


(a) sinA

2sin

B

2sin

C

2≤ 1

8;

(b) sin2 A

2+ sin2 B

2+ sin2 C

2≥ 3

4;

(c) cos2 A

2+ cos2 B

2+ cos2 C

2≤ 9

4;

(d) cosA

2cos

B

2cos

C

2≤ 3

√3

8;

(e) cscA

2+ csc

A

2+ csc

A

2≥ 6.

Solution: By Problem 8, we have

sinA

2sin

B

2sin

C

2≤ abc

(a + b)(b + c)(c + a).

The arithmetic–geometric means inequality yields

(a + b)(b + c)(c + a) ≥ (2√

ab)(2√

bc)(2√

ca) = 8abc.


Combining the last two equalities gives part (a).

Part (b) then follows from (a) and Problem 22. Part (c) then follows from part(b) by noting that 1 − sin2 x = cos2 x. Finally, by (c) and by the arithmetic–geometric means inequality, we have

9

4≥ cos2 A

2+ cos2 B

2+ cos2 C

2≥ 3 3

√cos2 A

2cos2 B

2cos2 C

2,

implying (d).

Again by Problem 8, we have

cscA

2≥ b + c

a= b

a+ c

a

and analogous formulas for csc B2 and csc C

2 . Then part (e) follows routinelyfrom the arithmetic–geometric means inequality.

Note: We present another approach to part (a). Note that sin A2 , sin B

2 , sin C2

are all positive. Let t = 3√

sin A2 sin B

2 sin C2 . It suffices to show that t ≤ 1

2 . Bythe arithmetic–geometric means inequality, we have

sin2 A

2+ sin2 B

2+ sin2 C

2≥ 3t2.

By Problem 22, we have 3t2 + 2t3 ≤ 1. Thus,

0 ≥ 2t3 + 3t2 − 1 = (t + 1)(2t2 + t − 1) = (t + 1)2(2t − 1).

Consequently, t ≤ 12 , establishing (a).


(a) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C;

(b) cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C;

(c) sin2 A + sin2 B + sin2 C = 2 + 2 cos A cos B cos C;

(d) cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1.

Conversely, if x, y, z are positive real numbers such that

x2 + y2 + z2 + 2xyz = 1,

show that there is an acute triangle ABC such that x = cos A, y = cos B,C = cos C.


Solution: Parts (c) and (d) follow immediately from (b) because cos 2x =1 − 2 sin2 x = 2 cos2 x − 1. Thus we show only (a) and (b).

(a) Applying the sum-to-product formulas and the fact that A + B + C =180◦, we find that

sin 2A + sin 2B + sin 2C = 2 sin(A + B) cos(A − B) + sin 2C

= 2 sin C cos(A − B) + 2 sin C cos C

= 2 sin C[cos(A − B) − cos(A + B)]= 2 sin C · [−2 sin A sin(−B)]= 4 sin A sin B sin C,

establishing (a).

(b) By the sum-to-product formulas, we have cos 2A+ cos 2B = 2 cos(A+B) cos(A − B) = −2 cos C cos(A − B), because A + B + C = 180◦.Note that cos 2C + 1 = 2 cos2 C. It suffices to show that

−2 cos C(cos(A − B) − cos C) = −4 cos A cos B cos C,

or cos C(cos(A − B) + cos(A + B)) = 2 cos A cos B cos C, which isevident by the sum-to-product formula cos(A − B) + cos(A + B) =2 cos A cos B.

From the given equality, we have 1 ≥ x2, 1 ≥ y2, and thus we may setx = cos A, y = cos B, where 0◦ ≤ A, B ≤ 90◦. Because x2 +y2 +z2 +2xyz

is an increasing function of z, there is at most one nonnegative value c suchthat the given equality holds. We know that one solution to this equality isz = cos C, where C = 180◦ −A−B. Because cos2 A+ cos2 B = x2 + y2 ≤1, we know that cos2 B ≤ sin2 A. Because 0◦ < A,B ≤ 90◦, we havecos B ≤ sin A = cos(90◦ − A), implying that A + B ≥ 90◦. Thus, C ≤ 90◦and cos C ≥ 0. Therefore, we must have z = cos C, as desired.

Note: Nevertheless, we present a cool proof of part (d). Consider the systemof equations

−x + (cos B)y + (cos C)z = 0

(cos B)x − y + (cos A)z = 0

(cos C)x + (cos A)y − z = 0.

Using the addition and subtraction formulas, one can easily see that(x, y, z) = (sin A, sin C, sin B) is a nontrivial solution. Hence the determinant


of the system is 0; that is,

0 =∣∣∣∣∣∣

−1 cos B cos C

cos B −1 cos A

cos C cos A −1

∣∣∣∣∣∣= −1 + 2 cos A cos B cos C + cos2 A + cos2 B + cos2 C,

as desired.


(a) 4R = abc

[ABC] ;

(b) 2R2 sin A sin B sin C = [ABC];(c) 2R sin A sin B sin C = r(sin A + sin B + sin C);

(d) r = 4R sinA

2sin

B

2sin

C

2;

(e) a cos A + b cos B + c cos C = abc

2R2 .

Solution: By the extended law of sines,

R = a

2 sin A= abc

2bc sin A= abc

4[ABC] ,

establishing (a). By the same token, we have

2R2 sin A sin B sin C = 1

2· (2R sin A)(2R sin B)(sin C)

= 1

2ab sin C = [ABC],

which is (b).

Note that2[ABC] = bc sin A = (a + b + c)r.

By the extended law of sines, we obtain

4R2 sin A sin B sin C = bc sin A = r(a + b + c)

= 2rR(sin A + sin B + sin C),

from which (c) follows.


By the law of cosines,

cos A = b2 + c2 − a2

2bc.

Hence, by the half-angle formulas, we have

sin2 A

2= 1 − cos A

2= 1

2− b2 + c2 − a2

4bc= a2 − (b2 + c2 − 2bc)

4bc

= a2 − (b − c)2

4bc= (a − b + c)(a + b − c)

4bc

= (2s − 2b)(2s − 2c)

4bc= (s − b)(s − c)

bc,

where 2s = a + b + c is the perimeter of triangle ABC. It follows that

sinA

2=√

(s − b)(s − c)

bc,

and the analogous formulas for sin B2 and sin C

2 . Hence

sinA

2sin

B

2sin

C

2= (s − a)(s − b)(s − c)

abc

= s(s − a)(s − b)(s − c)

sabc= [ABC]2

sabc

by Heron’s formula. It follows that

sinA

2sin

B

2sin

C

2= [ABC]

s· [ABC]

abc= r · 1

4R,

from which (d) follows.

Now we prove (e). By the extended law of sines, we have a cos A = 2R sin A ·cos A = R sin 2A. Likewise, b cos B = R sin 2B and c cos C = R sin 2C. By(a) and (b), we have

4R sin A sin B sin C = abc

2R2 .

It suffices to show that

sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C,

which is Problem 24(a).


26. Let s be the semiperimeter of triangle ABC. Prove that

(a) s = 4R cosA

2cos

B

2cos

C

2;

(b) s ≤ 3√

3

2R.

Solution: It is well known that rs = [ABC], or s = [ABC]r

. By Problem 25(b) and (d), part (a) follows from

s = R sin A sin B sin C

2 sin A2 sin B

2 sin C2

= 4R cosA

2cos

B

2cos

C

2

by the double-angle formulas.

We conclude part (b) from (a) and Problem 23 (d).


(a) cos A + cos B + cos C = 1 + 4 sinA

2sin

B

2sin

C

2;

(b) cos A + cos B + cos C ≤ 3

2.

Solution: By the sum-to-product and the double-angle formulas, we have

cos A + cos B = 2 cosA + B

2cos

A − B

2= 2 sin

C

2cos

A − B

2

and

1 − cos C = 2 sin2 C

2= 2 sin

C

2cos

A + B

2.


2 sinC

2

[cos

A − B

2− cos

A + B

2

]= 4 sin

A

2sin

B

2sin

C

2,

or,

cosA − B

2− cos

A + B

2= 2 sin

A

2sin

B

2,

which follows from the sum-to-product formulas, and hence (a) is established.

Recalling Problem 25 (c), we have

cos A + cos B + cos C = 1 + r

R. (∗)


Euler’s formula states that |OI |2 = R2 − 2Rr , where O and I are thecircumcenter and incenter of triangle ABC. Because |OI |2 ≥ 0, we haveR ≥ 2r , or r

R≤ 1

2 , from which (b) follows.

Note: Relation (∗) also has a geometric interpretation.

A

B CA1

B1C1

O

Figure 4.3.

As shown in the Figure 4.3, let O be the circumcenter, and let A1, B1, C1 be thefeet of the perpendiculars from O to sides BC, CA, AB, respectively. (ThusA1, B1, C1 are the midpoints of sides BC, CA, AB, respectively.) Because� AOB = 2C and triangle AOB is isosceles with |OA| = |OB| = R, wehave |OC1| = R cos C. Likewise, |OB1| = R cos B and |OA1| = R cos A.It suffices to show that

|OA1| + |OB1| + |OC1| = R + r.

Note that |OA| = |OB| = |OC| = R and |BA1| = |A1C|, |CB1| = |B1A|,|AC1| = |C1B|. Hence |AB| = 2|A1B1|, |BC| = 2|B1C1|, |CA| = 2|C1A1|.Let s denote the semiperimeter of triangle ABC.Applying Ptolemy’s theoremto cyclic quadrilaterals OA1CB1, OB1AC1, OC1BA1 yields

|A1B1| · |OC| = |A1C| · |OB1| + |CB1| · |OA1|,|B1C1| · |OA| = |B1A| · |OC1| + |AC1| · |OB1|,|C1A1| · |OB| = |C1B| · |OA1| + |BA1| · |OC1|.

Adding the above gives

Rs = |OA1|(s − |A1B|) + |OB1|(s − |B1C|) + |OC1|(s − |C1A|)= s(|OA1| + |OB1| + |OC1|) − [ABC]= s(|OA1| + |OB1| + |OC1|) − rs,


from which our desired result follows.


(a) cos A cos B cos C ≤ 1

8;

(b) sin A sin B sin C ≤ 3√

3

8;

(c) sin A + sin B + sin C ≤ 3√

3

2.

(d) cos2 A + cos2 B + cos2 C ≥ 3

4;

(e) sin2 A + sin2 B + sin2 C ≤ 9

4;

(f) cos 2A + cos 2B + cos 2C ≥ −3

2;

(g) sin 2A + sin 2B + sin 2C ≤ 3√

3

2.

Solution: For part (a), if triangle ABC is nonacute, the left-hand side of theinequality is nonpositive, and so the inequality is clearly true.

If ABC is acute, then cos A, cos B, cos C are all positive. To establish (a) and(d), we need only note that the relation between (a) and (d) and Problem 24(d)is similar to that of Problem 23(a) and (b) and Problem 22. (Please see thenote after the solution of Problem 23.)

The two inequalities in parts (d) and (e) are equivalent because cos2 x +sin2 x = 1.

By (e) and by the arithmetic–geometric means inequality, we have

9

4≥ sin2 A + sin2 B + sin2 C ≥ 3

3√

sin2 A sin2 B sin2 C,


From (a−b)2+(b−c)2+(c−a)2 ≥ 0 or by application of Cauchy–Schwarzinequality, we can show that 3

(a2 + b2 + c2

) ≥ (a + b + c)2. By (e) and bysetting a = sin A, b = sin B, c = sin C, we obtain (c).

Part (f) follows from (e) and cos 2x = 2 cos2 x − 1. Finally, (g) follows from(b) and the identity

sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C


proved in Problem 25(e).

29. Prove thattan 3x

tan x= tan

(π3

− x)

tan(π

3+ x)


Solution: From the triple-angle formulas, we have

tan 3x = 3 tan x − tan3 x

1 − 3 tan2 x

= tan x · (√

3 − tan x)(√

3 + tan x)

(1 − √3 tan x)(1 + √

3 tan x)

=√

3 − tan x

1 + √3 tan x

· tan x ·√

3 + tan x

1 − √3 tan x

= tan(π

3− x)

tan x tan(π

3+ x)


30. [AMC12P 2002] Given that

(1 + tan 1◦)(1 + tan 2◦) · · · (1 + tan 45◦) = 2n,

find n.

First Solution: Note that

1 + tan k◦ = 1 + sin k◦

cos k◦ = cos k◦ + sin k◦

cos k◦

=√

2 sin(45 + k)◦

cos k◦ =√

2 cos(45 − k)◦

cos k◦ .

Hence

(1 + tan k◦)(1 + tan(45 − k)◦) =√

2 cos(45 − k)◦

cos k◦ ·√

2 cos(k)◦

cos(45 − k)◦= 2.

It follows that

(1 + tan 1◦)(1 + tan 2◦) · · · (1 + tan 45◦)= (1 + tan 1◦)(1 + tan 44◦)(1 + tan 2◦)(1 + tan 43◦)

· · · (1 + tan 22◦)(1 + tan 23◦)(1 + tan 45◦)= 223,


implying that n = 23.

Second Solution: Note that

(1 + tan k◦)(1 + tan(45 − k)◦)= 1 + [tan k◦ + tan(45 − k)◦] + tan k◦ tan(45 − k)◦

= 1 + tan 45◦[1 − tan k◦ tan(45 − k)◦] + tan k◦ tan(45 − k)◦

= 2.

Hence

(1 + tan 1◦)(1 + tan 2◦) · · · (1 + tan 45◦)= (1 + tan 1◦)(1 + tan 44◦)(1 + tan 2◦)(1 + tan 43◦)

· · · (1 + tan 22◦)(1 + tan 23◦)(1 + tan 45◦)= 223,

implying that n = 23.

31. [AIME 2003] Let A = (0, 0) and B = (b, 2) be points in the coordinate plane.Let ABCDEF be a convex equilateral hexagon such that � FAB = 120◦,AB ‖ DE, BC ‖ EF , and CD ‖ FA, and the y coordinates of its verticesare distinct elements of the set {0, 2, 4, 6, 8, 10}. The area of the hexagon canbe written in the form m

√n, where m and n are positive integers and n is not

divisible by the square of any prime. Find m + n.

Note: Without loss of generality, we assume that b > 0. (Otherwise, we canreflect the hexagon across the y axis.) Let the x coordinates of C, D, E, and F

be c, d, e, and f , respectively. Note that the y coordinate of C is not 4, sinceif it were, the fact |AB| = |BC| would imply that A, B, and C are collinear orthat c = 0, implying that ABCDEF is concave. Therefore, F = (f, 4). Since−→AF = −→

CD, C = (c, 6) and D = (d, 10), and so E = (e, 8). Because the y

coordinates of B, C, and D are 2, 6, and 10, respectively, and |BC| = |CD|,we conclude that b = d. Since

−→AB = −→

ED, e = 0. Let a denote the side lengthof the hexagon. Then f < 0. We need to compute

[ABCDEF ] = [ABDE] + [AEF ] + [BCD] = [ABDE] + 2[AEF ]= b · AE + (−f ) · AE = 8(b − f ).


Solution:

A

B

C

D

E

F

Figure 4.4.

First Solution: Note that f 2 + 16 = |AF |2 = a2 = |AB|2 = b2 + 4. Applythe law of cosines in triangle ABF to obtain 3a2 = |BF |2 = (b − f )2 + 4.We have three independent equations in three variables. Hence we can solvethis system of equations. The quickest way is to note that

b2 + f 2 − 2bf + 4 = (b − f )2 + 4 = 3a2 = a2 + b2 + 4 + f 2 + 16,

implying that a2 + 16 = −2bf . Squaring both sides gives

a4 + 32a2 + 162 = 4b2f 2 = 4(a2 − 4)(a2 − 16) = 4a4 − 80a2 + 162,

or 3a4 − 112a2 = 0. Hence a2 = 1123 , and so b = 10√

3and f = − 8√

3.

Therefore, [ABCDEF ] = 8(b − f ) = 48√

3, and the answer to the problemis 51.

Second Solution: Let α denote the measure (in degrees) of the standard angleformed by the line AB and and the x axis. Then the standard angle formed bythe line AF and the x axis is β = 120◦ + α. By considering the y coordinatesof B and F , we have a sin α = 2 and

4 = a sin(120◦ + α) = a√

3 cos α

2− a sin α

2= a

√3 cos α

2− 1,

by the addition and subtraction formulas. Hence a cos α = 10√3

. Thus, by

considering the x coordinates of B and F , we have b = a cos α = 10√3

and

f = a cos(120◦ + α) = −a cos α

2− a

√3 sin α

2= − 8√

3.


It follows that [ABCDEF ] = 48√

3.

Note: The vertices of the hexagon are A = (0, 0), B =(

10√3, 2)

, C =(6√

3, 6)

, D =(

10√3, 10)

, E = (0, 8), and F =(− 8√

3, 4)

.

32. Show that one can use a composition of trigonometry buttons such as, sin, cos,tan, sin−1, cos−1, and tan−1, to replace the broken reciprocal button on acalculator.

Solution: Because cos−1 sin θ = π/2 − θ , and tan(π/2 − θ) = 1/ tan θ for0 < θ < π/2, we have for any x > 0,

tan cos−1 sin tan−1 x = tan(π

2− tan−1 x

)= 1

x,

as desired. It is not difficult to check that tan sin−1 cos tan−1 will also do thetrick.

33. In triangle ABC, A − B = 120◦ and R = 8r . Find cos C.

Solution: From Problem 25(d), it follows that

2 sinA

2sin

B

2sin

C

2= 1

16,

or (cos

A − B

2− cos

A + B

2

)sin

C

2= 1

16,

by the product-to-sum formulas. Taking into account that A − B = 120◦,we obtain (

1

2− sin

C

2

)sin

C

2= 1

16,

or (1

4− sin

C

2

)2

= 0,

yielding sin C2 = 1

4 . Hence cos C = 1 − 2 sin2 C2 = 7

8 .


34. Prove that in a triangle ABC,

a − b

a + b= tan

A − B

2tan

C

2.

Solution: From the law of sines and the sum-to-product formulas, we have

a − b

a + b= sin A − sin B

sin A + sin B= 2 sin A−B

2 cos A+B2

2 sin A+B2 cos A−B

2

= tanA − B

2cot

A + B

2= tan

A − B

2tan

C

2,

as desired.

35. In triangle ABC, ab

= 2 + √3 and � C = 60◦. Find the measure of angles A

and B.

Solution: From the previous problem we deduce that

ab

− 1ab

+ 1= tan

A − B

2tan

C

2.

It follows that1 + √

3√3 + 3

= tanA − B

2· 1√

3,

and so tan A−B2 = 1. Thus A−B = 90◦, and since A+B = 180◦−C = 120◦,

we obtain A = 105◦ and B = 15◦.

36. Let a, b, c be real numbers, all different from −1 and 1, such that a + b + c =abc. Prove that

a

1 − a2 + b

1 − b2 + c

1 − c2 = 4abc

(1 − a2)(1 − b2)(1 − c2).

Solution: Let a = tan x, b = tan y, c = tan z, where x, y, z �= kπ4 , for all

integers k. The condition a + b + c = abc translates to tan(x + y + z) = 0,as indicated in notes after Problem 20(a). From the double-angle formulas,it follows that

tan(2x + 2y + 2z) = 2 tan(x + y + z)

1 − tan2(x + y + z)= 0.


Hencetan 2x + tan 2y + tan 2z = tan 2x tan 2y tan 2z,

using a similar argument to the one in Problem 20(a). This implies that

2 tan x

1 − tan2 x+ 2 tan y

1 − tan2 y+ 2 tan z

1 − tan2 z

= 2 tan x

1 − tan2 x· 2 tan y

1 − tan2 y· 2 tan z

1 − tan2 z,

and the conclusion follows.

37. Prove that a triangle ABC is isosceles if and only if

a cos B + b cos C + c cos A = a + b + c

2.

Solution: By the extended law of sines, a = 2R sin A, b = 2R sin B, andc = 2R sin C. The desired identity is equivalent to

2 sin A cos B + 2 sin B cos C + 2 sin C cos A = sin A + sin B + sin C,

orsin(A + B) + sin(A − B) + sin(B + C)

+ sin(B − C) + sin(C + A) + sin(C − A)

= sin A + sin B + sin C.

Because A + B + C = 180◦, sin(A + B) = sin C, sin(B + C) = sin A,sin(C + A) = sin B. The last equality simplifies to

sin(A − B) + sin(B − C) + sin(C − A) = 0,

which in turn is equivalent to

4 sinA − B

2sin

B − C

2sin

C − A

2= 0,

by Problem 15. The conclusion now follows.

38. Evaluatecos a cos 2a cos 3a · · · cos 999a,

where a = 2π1999 .


Solution: Let P denote the desired product, and let

Q = sin a sin 2a sin 3a · · · sin 999a.

Then

2999PQ = (2 sin a cos a)(2 sin 2a cos 2a) · · · (2 sin 999a cos 999a)

= sin 2a sin 4a · · · sin 1998a

= (sin 2a sin 4a · · · sin 998a)[− sin(2π − 1000a)]· [− sin(2π − 1002a)] · · · [− sin(2π − 1998a)]

= sin 2a sin 4a · · · sin 998a sin 999a sin 997a · · · sin a = Q.

It is easy to see that Q �= 0. Hence the desired product is P = 12999 .

39. Determine the minimum value of

sec4 α

tan2 β+ sec4 β

tan2 α

over all α, β �= kπ2 , where k is in Z.

Solution: Set a = tan2 α and b = tan2 β. It suffices to determine the minimumvalue of

(a + 1)2

b+ (b + 1)2

a,

with a, b ≥ 0. We have

(a + 1)2

b+ (b + 1)2

a= a2 + 2a + 1

b+ b2 + 2b + 1

a

=(

a2

b+ 1

b+ b2

a+ 1

a

)+ 2

(a

b+ b

a

)

≥ 44

√a2

b· 1

b· b2

a· 1

a+ 4

√a

b· b

a= 8,

by the arithmetic–geometric means inequality. Equality holds when a =b = 1; that is, α = ±45◦ + k · 180◦ and β = ±45◦ + k · 180◦, for integers k.


40. Find all pairs (x, y) of real numbers with 0 < x < π2 such that

(sin x)2y

(cos x)y2/2

+ (cos x)2y

(sin x)y2/2

= sin 2x.

Solution: The arithmetic–geometric means inequality gives

(sin x)2y

(cos x)y2/2

+ (cos x)2y

(sin x)y2/2

≥ 2(sin x cos x)y−y2/4.

It follows that

2 sin x cos x = sin 2x ≥ 2(sin x cos x)y−y2/4,

and because sin x cos x < 1, it follows that 1 ≤ y − y2/4, or (1 − y/2)2 ≤ 0.It follows that all the equalities hold; that is, y = 2 and sin x = cos x, and sothere is a unique solution: (x, y) = (π4 , 2

).

41. Prove that cos 1◦ is an irrational number.

Solution: Assume, for the sake of contradiction, that cos 1◦ is rational. Thenso is cos 2◦ = 2 cos2 1◦ − 1. Using the identity

cos(n◦ + 1◦) + cos(n◦ − 1◦) = 2 cos n◦ cos 1◦, (∗)

we obtain by strong induction that cos n◦ is rational for all integers n ≥ 1. Butthis is clearly false, because, for example, cos 30◦ is not rational, yielding acontradiction.

Note: For the reader not familiar with the idea of induction. We can reasonin the following way. Under the assumption that both cos 1◦ and cos 2◦ arerational, relation (∗) implies that cos 3◦ is rational, by setting n = 2 in therelation (∗). Similarly, by the assumption that both cos 2◦ and cos 3◦ are ratio-nal, relation (∗) implies that cos 4◦ is rational, by setting n = 4 in the relation(∗). And so on. We conclude that cos n◦ is rational, for all positive integers n,under the assumption that cos 1◦ is rational.

42. [USAMO 2002 proposal, by Cecil Rousseau] Find the maximum value of

S = (1 − x1)(1 − y1) + (1 − x2)(1 − y2)


if x21 + x2

2 = y21 + y2

2 = c2.

Solution: If we interpret x1 and x2 are the coordinates of a point; that is,assume that P = (x1, x2), then P lies on a circle centered at the origin withradius c. We can describe the circle parametrically; that is, write x1 = c cos θ ,x2 = c sin θ , and similarly, y1 = c cos φ, y2 = c sin φ. Then

S = 2 − c(cos θ + sin θ + cos φ + sin φ) + c2(cos θ cos φ + sin θ sin φ)

= 2 − √2c[sin(θ + π/4) + sin(φ + π/4)] + c2 cos(θ − φ)

≤ 2 + 2√

2c + c2 = (√

2 + c)2,

with equality at θ = φ = 5π/4, that is, x1 = x2 = y1 = y2 = −c√

22 .

43. Prove thatsin3 a

sin b+ cos3 a

cos b≥ sec(a − b)

for all 0 < a, b < π2 .

Solution: Multiplying the two sides of the inequality by sin a sin b+cos a cos b =cos(a − b), we obtain the equivalent form(

sin3 a

sin b+ cos3 a

cos b

)(sin a sin b + cos a cos b) ≥ 1.

But this follows from Cauchy–Schwarz inequality because according to thisinequality, the left-hand side is greater than or equal to (sin2 a +cos2 a)2 = 1.

44. If sin α cos β = − 12 , what are the possible values of cos α sin β?

Solution: Note that

sin(α + β) = sin α cos β + cos α sin β = −1

2+ cos α sin β.

Because −1 < sin(α + β) ≤ 1, it follows that − 12 ≤ cos α sin β < 3

2 .Similarly, because sin(α − β) = sin α cos β − cos α sin β, we conclude that− 3

2 ≤ cos α sin β < 12 . Combining the above results shows that

−1

2≤ cos α sin β ≤ 1

2.


But we have not shown that indeed, cos α sin β can obtain all values in theinterval

[− 12 , 1

2

]. To do this, we consider

(cos α sin β)2 = (1 − sin2 α)(1 − cos2 β)

= 1 − (sin2 α + cos2 β) + sin2 α cos2 β

= 5

4− (sin2 α + cos2 β)

= 5

4− (sin α + cos β)2 + 2 sin α cos β

= 1

4− (sin α + cos β)2.

Let x = sin α and y = cos β. Then −1 ≤ x, y,≤ 1 and xy = − 12 . Consider

the range of the sum s = sin α + cos β = x + y. If xy = − 12 and x + y = s,

then x and y are the roots of the quadratic equation

u2 − su − 1

2= 0. (∗)

Thus, {x, y} ={

s+√s2+22 , s−√

s2+22

}. By checking the boundary condition

s+√s2+22 ≤ 1, we obtain s ≤ 1

2 . By checking similar boundary conditions, weconclude that the equation (∗) has a pair of solutions x and y with −1 ≤ x, y ≤1 for all − 1

2 ≤ s ≤ 12 . Because both the sine and cosine functions are surjective

functions from R to the interval [−1, 1], the range of s = sin α + cos β is[− 12 , 1

2

]for sin α cos β = − 1

2 . Thus, the range of s2 is[0, 1

2

]. Thus the range

of (cos α sin β)2 is[0, 1

4

], and so the range of cos α sin β is

[− 12 , 1

2

].

45. Let a, b, c be real numbers. Prove that

(ab + bc + ca − 1)2 ≤ (a2 + 1)(b2 + 1)(c2 + 1).

Solution: Let a = tan x, b = tan y, c = tan z with −π2 < x, y, z < π

2 .Then a2 + 1 = sec2 x, b2 + 1 = sec2 y, and c2 + 1 = sec2 z. Multiplying bycos2 x cos2 y cos2 z on both sides of the desired inequality gives

[(ab + bc + ca − 1) cos x cos y cos z]2 ≤ 1.

Note that

(ab + bc) cos x cos y cos z = sin x sin y cos z + sin y sin z cos x

= sin y sin(x + z)


and

(ca − 1) cos x cos y cos z = sin z sin x cos y − cos x cos y cos z

= − cos y cos(x + z).

Consequently, we obtain

[(ab + bc + ca − 1) cos x cos y cos z]2

= [sin y sin(x + z) − cos y cos(x + z)]2

= cos2(x + y + z) ≤ 1,

as desired.

46. Prove that

(sin x + a cos x)(sin x + b cos x) ≤ 1 +(

a + b

2

)2

.

Solution: If cos x = 0, the desired inequality reduces to sin2 x ≤ 1+( a+b2

)2,

which is clearly true. We assume that cos x �= 0. Dividing both sides of thedesired inequality by cos2 x gives

(tan x + a)(tan x + b) ≤[

1 +(

a + b

2

)2]

sec2 x.

Set t = tan x. Then sec2 x = 1 + t2. The above inequality reduces to

t2 + (a + b)t + ab ≤(

a + b

2

)2

t2 + t2 +(

a + b

2

)2

+ 1,

or (a + b

2

)2

t2 + 1 − (a + b)t +(

a + b

2

)2

− ab ≥ 0.

The last inequality is equivalent to((a + b)t

2− 1

)2

+(

a − b

2

)2

≥ 0,

which is evident.


47. Prove that

| sin a1| + | sin a2| + · · · + | sin an| + | cos(a1 + a2 + · · · + an)| ≥ 1.

Solution: We proceed by induction on n. The base case holds, because

| sin a1| + | cos a1| ≥ sin2 a1 + cos2 a1 = 1.

For the inductive step, in order to prove that

| sin a1| + | sin a2| + · · · + | sin an+1| + | cos(a1 + a2 + · · · + an+1)| ≥ 1,

it suffices to show that

| sin an+1| + | cos(a1 + a2 + · · · + an+1)| ≥ | cos(a1 + a2 + · · · + an)|for all real numbers a1, a2, . . . , an+1. Let sk = a1 + a2 + · · · + ak , for k =1, 2, . . . , n+1. The last inequality becomes | sin an+1|+| cos sn+1| ≥ | cos sn|.Indeed, by the addition and subtraction formulas, we have

| cos sn| = | cos(sn+1 − an+1)|= | cos sn+1 cos an+1 + sin sn+1 sin an+1|= | cos sn+1 cos an+1| + | sin sn+1 sin an+1|≤ | cos sn+1| + | sin an+1|,

as desired.

48. [Russia 2003, by Nazar Agakhanov] Find all angles α for which the three-element set

S = {sin α, sin 2α, sin 3α}is equal to the set

T = {cos α, cos 2α, cos 3α}.

Solution: The answers are α = π8 + kπ

2 for all integers k.

Because S = T , the sums of the elements in S and T are equal to each other;that is,

sin α + sin 2α + sin 3α = cos α + cos 2α + cos 3α.

Applying the sum-to-product formulas to the first and the third summandson each side of the last equation gives

2 sin 2α cos α + sin 2α = 2 cos 2α cos α + cos 2α,


orsin 2α(2 cos α + 1) = cos 2α(2 cos α + 1).

If 2 cos α + 1 = 0, then cos α = − 12 , and so α = ± 2π

3 + 2kπ for all integersk. It is then not difficult to check that S �= T and both of S and T are notthree-element sets.

It follows that 2 cos α+1 �= 0, implying that sin 2α = cos 2α; that is, tan 2α =1. The possible answers are α = π

8 + kπ2 for all integers k. Because π

8 + 3π8 =

π2 , cos π

8 = sin 3π8 . It not difficult to check that all such angles satisfy the

conditions of the problem.

49. Let {Tn(x)}∞n=0 be the sequence of polynomials such that T0(x) = 1, T1(x) =x, Ti+1 = 2xTi(x)−Ti−1(x) for all positive integers i. The polynomial Tn(x)

is called the nth Chebyshev polynomial.

(a) Prove that T2n+1(x) and T2n(x) are odd and even functions, respectively;

(b) Prove that Tn+1(x) > Tn(x) > 1 for real numbers x with x > 1;

(c) Prove that Tn(cos θ) = cos(nθ) for all nonnegative integers n;

(d) Determine all the roots of Tn(x);

(e) Determine all the roots of Pn(x) = Tn(x) − 1.

Solution: Parts (a) and (b) are simple facts that will be useful in establishing(e). We present them together.

(a) We apply strong induction on n. Note that T0 = 1 and T1 = x areeven and odd, respectively. Assume that T2n−1 and T2n are odd and even,respectively. Then 2xT2n is odd, and so T2n+1 = 2xT2n − T2n−1 is odd.Thus 2xT2n+1 is even, and so T2n+2 = 2xT2n+1 − T2n is even. Thiscompletes our induction.

(b) We apply strong induction on n. For n = 0, T1(x) = x > 1 = T0(x) forx > 1. Assume that Tn+1(x) > Tn(x) > 1 for x > 1 and n ≤ k, wherek is some nonnegative integer. For n = k + 1, the induction hypothesisyields

Tk+2(x) = 2xTk+1(x) − Tk(x) > 2Tk+1(x) − Tk(x)

= Tk+1(x) + Tk+1(x) − Tk(x) > Tk+1(x),

completing our induction.


(c) We again apply strong induction on n. The base cases for n = 0 andn = 1 are trivial. Assume that Tn(cos θ) = cos(nθ) for n ≤ k, where k

is some positive integer. The induction hypothesis gives

Tk+1(cos θ) = 2 cos θTk(cos θ) − Tk−1(cos θ)

= 2 cos θ cos kθ − cos[(k − 1)θ ].

By the product-to-sum formulas, we have

2 cos θ cos kθ = cos[(k + 1)θ ] + cos[(k − 1)θ ].

It follows that Tk+1(cos θ) = cos[(k + 1)θ ], completing our induction.

(d) It is clear that Tn is a polynomial of degree n, and so it has at most n realroots. Note that y = cos x is a one-to-one and onto mapping from theinterval

[0, π

2

]. By (c), we conclude that Tn has exactly n distinct real

roots, and they form the set

S ={

coskπ

2n, k = 1, 3, . . . , 2n − 1

}.

(e) By (a), Tn is either even or odd, and so by (b), |Tn(x)| > 1 for x < −1.Thus, all the roots of Pn lie in the interval [−1, 1]. We consider two cases.

Assume first that n is even. A real number is a root of Pn if and only ifit is in the set

Se ={

coskπ

n, k = 0, 2, . . . , n

}.

Assume next that n is odd. A real number is a root of Pn if and only if itis in the set

Se ={

coskπ

n, k = 0, 2, . . . , n − 1

}.

50. [Canada 1998] Let ABC be a triangle with � BAC = 40◦ and � ABC = 60◦.Let D and E be the points lying on the sides AC and AB, respectively, suchthat � CBD = 40◦ and � BCE = 70◦. Segments BD and CE meet at F .Show that AF ⊥ BC.


Solution:

A

B

CD

EF G

Figure 4.5.

Note that � ABD = 20◦, � BCA = 80◦, and � ACE = 10◦. Let G be thefoot of the altitude from A to BC. Then � BAG = 90◦ − � ABC = 30◦ and� CAG = 90◦ − � BCA = 10◦. Now,

sin � BAG sin � ACE sin � CBD

sin � CAG sin � BCE sin � ABD= sin 30◦ sin 10◦ sin 40◦

sin 10◦ sin 70◦ sin 20◦

=12 (sin 10◦)(2 sin 20◦ cos 20◦)

sin 10◦ cos 20◦ sin 20◦= 1.

Then by the trigonometric form of Ceva’s theorem, lines AG, BD, and CE areconcurrent. Therefore, F lies on segment AG, and so line AF is perpendicularto the line BC, as desired.

51. [IMO 1991] Let S be an interior point of triangle ABC. Show that at least oneof � SAB, � SBC, and � SCA is less than or equal to 30◦.

First Solution: The given conditions in the problem motivate us to considerthe Brocard point P of triangle ABC with α = � PAB = � PBC = � PCA.Because S (see Figure 4.6) lies inside or on the boundary of at least one of thetriangles PAB, PBC, and PCA, at least one of � SAB, � SBC, and � SCA

is less than or equal to α. It suffices to show that α ≤ 30◦; that is, sin α ≤ 12

or csc2 α ≥ 4, by considering the range of α.

We have shown that

csc2 α = csc2 A + csc2 B + csc2 C.


By Problem 28(e) and Cauchy–Schwarz inequality, we have

9

4csc2 α ≥

(sin2 A + sin2 B + sin2 C

) (csc2 A + csc2 B + csc2 C

)≥ 9,

implying that csc2 α ≥ 4, as desired.

AB

C

P

Figure 4.6.

Second Solution: We use radian measure in this solution; that is, we wantto show that one of � SAB, � SBC, and � SCA is less than or equal to π

6 .Set x = � SAB, y = � SBC, and z = � SCA. Let da, db and dc denote thedistance from S to sides BC, CA, and AB. Then

dc = SA sin x = SB sin(B − y),

da = SB sin y = SC sin(C − z),

db = SC sin z = SA sin(A − x).

Multiplying the last three equations together gives

sin x sin y sin z = sin(A − x) sin(B − y) sin(C − z). (∗)

If x + y + z ≤ π2 , then the conclusion of the problem is clearly true. Now we

assume that x + y + z > π2 ; that is, (A − x) + (B − y) + (C − z) < π

2 .

Now we consider the function f (x) = ln(sin x), where 0 < x < π2 . Then

the first derivative of f (x) is f ′(x) = cos xsin x

= cot x, and the second deriva-tive is f ′′(x) = − csc2 x < 0. Hence f (x) is concave down. By Jensen’sinequality, we have

1

3(ln sin(A − x) + ln sin(B − y) + ln sin(C − z))

≤ ln sin(A − x) + (B − y) + (C − z)

3,


implying that

ln (sin(A − x) sin(B − y) sin(C − z))13 ≤ ln sin

6

π= ln

1

2,

or sin(A−x) sin(B −y) sin(C − z) ≤ 18 . Thus sin x sin y sin z ≤ 1

8 , implyingthat at least one of sin x, sin y, and sin z is less than or equal to 1

2 , as desired.

Third Solution: We also have a clever way to apply equation (∗) withoutusing Jensen’s inequality. From equation (∗) we have

(sin x sin y sin z)2 = sin x sin(A − x) sin y sin(B − y) sin z sin(C − z).

Applying the product-to-sum formulas and the double-angle formulas gives2 sin x sin(A − x) = cos(A − 2x) − cos A ≤ 1 − cos A = 2 sin2 A

2 , orsin x sin(A − x) ≤ sin2 A

2 and its analogous forms. (This step can also becarried by applying Jensen’s inequality. The reader might want to do so as anexercise.) It follows, by Problem 23(a), that

sin x sin y sin z ≤ sinA

2sin

B

2sin

C

2≤ 1

8,

from which our desired result follows.

52. Let a = π7 .

(a) Show that sin2 3a − sin2 a = sin 2a sin 3a;

(b) Show that csc a = csc 2a + csc 4a;

(c) Evaluate cos a − cos 2a + cos 3a;

(d) Prove that cos a is a root of the equation 8x3 + 4x2 − 4x − 1 = 0;

(e) Prove that cos a is irrational;

(f) Evaluate tan a tan 2a tan 3a;

(g) Evaluate tan2 a + tan2 2a + tan2 3a;

(h) Evaluate tan2 a tan2 2a + tan2 2a tan2 3a + tan2 3a tan2 a.

(i) Evaluate cot2 a + cot2 2a + cot2 3a.

Solution: Many of the desired results are closely related. Parts (d) and (e)will be presented together, as will (f), (g), (h), and (i).


(a) By the sum-to-product, the difference-to-product and the double-angle formulas, we have

sin2 3a − sin2 a = (sin 3a + sin a)(sin 3a − sin a)

= (2 sin 2a cos a)(2 sin a cos 2a)

= (2 sin 2a cos 2a)(2 sin a cos a)

= sin 4a sin 2a = sin 2a sin 3a,

as desired. The last identity is evident by noting that 4a + 3a = π (andso sin 3a = sin 4a).

(b) It suffices to show that

sin 2a sin 4a = sin a(sin 2a + sin 4a),

or2 sin a cos a sin 4a = sin a(2 sin 3a cos a),

by the sum-to-product formulas.

(c) The answer is 12 . It suffices to show that cos 2a + cos 4a + cos 6a = − 1

2 .This is a special case (n = 3) of a more general result:

t = cos 2x + cos 4x + · · · + cos 2nx = −1

2,

where x = π2n+1 . Indeed, applying the product-to-sum formulas gives

2 sin x cos kx = sin(k + 1)x − sin(k − 1)x, and so

2t sin x = 2 sin x(cos 2x + cos 4x + · · · + cos 2nx)

= [sin 3x − sin x] + [sin 5x − sin 3x]+ · · · + [sin(2n + 1)x − sin(2n − 1)x]

= sin(2n + 1)x − sin x = − sin x,

from which the desired equality follows.

(d) Because 3a + 4a = π , it follows that sin 3a = sin 4a. The double-angleand triple-angle formulas yield

sin a(3 − 4 sin2 a) = 2 sin 2a cos 2a = 4 sin a cos a cos 2a,

or 3 − 4(1 − cos2 a) = 4 cos a(2 cos2 a − 1). It follows that

8 cos3 a − 4 cos2 a − 4 cos a + 1 = 0,

establishing (c). Thus u = 2 cos a is the root of the cubic equation

u3 − u2 − 2u + 1 = 0. (∗)


By Gauss’s lemma, the only possible rational roots of the above cubicequation are 1 and −1. It is easy to see that neither is a root. Hence theabove equation has no rational root, implying that 2 cos a is not rational.Therefore, cos a is not rational.

Note: Although converting to equation (∗) is not necessary, it is a veryeffective technique. Instead of checking of eight possible rational rootsfrom the set

{± 18 , ± 1

4 , ± 12 , ±1

}of the equation

8x3 − 4x2 − 4x + 1 = 0,

we need to check only two possibilities for equation (∗).

(f) Because 3a + 4a = π , it follows that tan 3a + tan 4a = 0. The double-angle and the addition and subtraction formulas yield

tan a + tan 2a

1 − tan a tan 2a+ 2 tan 2a

1 − tan2 2a= 0,

ortan a + 3 tan 2a − 3 tan a tan2 2a − tan3 2a = 0.

Set tan a = x. Then tan 2a = 2 tan a1−tan2 a

= 2x1−x2 . Hence

x + 6x

1 − x2 − 12x3

(1 − x2)2 − 8x3

(1 − x2)3 = 0,

or (1 − x2

)3 + 6(

1 − x2)2 − 12x2

(1 − x2

)− 8x2 = 0.

Expanding the left-hand side of the above equation gives

x6 − 21x4 + 35x2 − 7 = 0. (†)

Thus tan a is a root of the above equation. Note that 6a + 8a = 2π and9a + 12a = 3π , and so tan[3(2a)] + tan[4(2a)] = 0 and tan[3(3a)] +tan[4(3a)] = 0. Hence tan 2a and tan 3a are the also the roots of equation(†). Therefore, tan2 ka, k = 1, 2, 3, are the distinct roots of the cubicequation

x3 − 21x2 + 35x − 7 = 0.

By Viète’s theorem, we have

tan2 a + tan2 2a + tan2 3a = 21;tan2 a tan2 2a + tan2 2a tan2 3a + tan2 3a tan2 a = 35;

tan2 a tan2 2a tan2 3a = 7.


Thus the answers for (f), (g), (h), and (i) are√

7, 21, 35, and 5, respec-tively.

Note: It is not difficult to check that the roots of the equation (†) are tan π7 ,

tan 2π7 , …, tan 6π

7 . On the other hand, it is interesting to note that 1, −21, 35, −7,

the coefficients of equation (†) are(7

0

), −(72), (74), −(76). In general, we have the

following result: For positive integers n, let an = π2n+1 . Then sin(2n+1)an =

0. The expansion formulas give

0 = sin(2n + 1)an

=(

2n + 1

1

)cos2n an sin an −

(2n + 1

3

)cos2n−2 an sin3 an

+(

2n + 1

5

)cos2n−4 an sin5 an − · · ·

= cos2n+1 an

[(2n + 1

1

)tan an −

(2n + 1

3

)tan3 an

+(

2n + 1

5

)tan5 an − · · ·

].

Because cos an �= 0, it follows that(2n + 1

1

)tan an −

(2n + 1

3

)tan3 an +

(2n + 1

5

)tan5 an − · · · = 0;

that is, tan an is a root of the equation(2n + 1

1

)x −(

2n + 1

3

)x3 + · · · + (−1)n

(2n + 1

2n + 1

)x2n+1 = 0,

or (2n + 1

0

)x2n −

(2n + 1

2

)x2n−2 + · · · + (−1)n

(2n + 1

2n

)= 0.

It is not difficult to see that the roots of the above equation are tan π2n+1 ,

tan 2π2n+1 , . . . , tan 2nπ

2n+1 . It is also not difficult to see that the roots of theequation(

2n + 1

0

)xn −

(2n + 1

2

)xn−1 + · · · + (−1)n

(2n + 1

2n

)= 0

are tan2 π2n+1 , tan2 2π

2n+1 , . . . , tan2 nπ2n+1 . By Viète’s theorem, we can obtain

more general results, such asn∑

k=1

cot2 kπ

2n + 1=(2n+1

2n−2

)(2n+12n

) = n(2n − 1)

3.

5Solutions to Advanced Problems

1. Two exercises on sin k◦ sin(k + 1)◦:

(a) [AIME2 2000] Find the smallest positive integer n such that

1

sin 45◦ sin 46◦ + 1

sin 47◦ sin 48◦ + · · · + 1

sin 133◦ sin 134◦

= 1

sin n◦ .

(b) Prove that

1

sin 1◦ sin 2◦ + 1

sin 2◦ sin 3◦ + · · · + 1

sin 89◦ sin 90◦

= cos 1◦

sin2 1◦ .

Solution: Note that

sin 1◦ = sin[(x + 1)◦ − x◦]= sin(x + 1)◦ cos x◦ − cos(x + 1)◦ sin x◦.

Thussin 1◦

sin x◦ sin(x + 1)◦= cos x◦ sin(x + 1)◦ − sin x◦ cos(x + 1)◦

sin x◦ sin(x + 1)◦= cot x◦ − cot(x + 1)◦.


(a) Multiplying both sides of the given equation by sin 1◦, we have

sin 1◦

sin n◦ = (cot 45◦ − cot 46◦) + (cot 47◦ − cot 48◦)

+ · · · + (cot 133◦ − cot 134◦)= cot 45◦ − (cot 46◦ + cot 134◦) + (cot 47◦ + cot 133◦)

− · · · + (cot 89◦ + cot 91◦) − cot 90◦

= 1.

Therefore, sin n◦ = sin 1◦, and the least possible integer value for n is 1.

(b) The left-hand side of the desired equation is equal to

89∑k=1

1

sin k◦ sin(k + 1)◦= 1

sin 1◦89∑

k=1

[cot k◦ − cot(k + 1)◦

]= 1

sin 1◦ · cot 1◦ = cos 1◦

sin2 1◦ ,

thus completing the proof.

2. [China 2001, by Xiaoyang Su] Let ABC be a triangle, and let x be a nonneg-ative real number. Prove that

ax cos A + bx cos B + cx cos C ≤ 1

2(ax + bx + cx).

Solution: By symmetry, we may assume that a ≥ b ≥ c. Hence A ≥ B ≥ C,and so cos A ≤ cos B ≤ cos C. Thus

(ax − bx)(cos A − cos B) ≤ 0,

orax cos A + bx cos B ≤ ax cos B + bx cos A.

Adding the last inequality with its analogous cyclic symmetric forms and thenadding ax cos A+bx cos B +cx cos C to both sides of the resulting inequalitygives

3(ax cos A + bx cos B + cx cos C)

≤ (ax + bx + cx)(cos A + cos B + cos C),

5. Solutions to Advanced Problems 127

from which the desired result follows as a consequence of Introductory Prob-lem 27(b).

Note: The above solution is similar to the proof of Chebyshev’s inequal-ity. We can also apply the rearrangement inequality to simplify our work.Because a ≥ b ≥ c and cos A ≤ cos B ≤ cos C, we have

ax cos A + bx cos B + cx cos C ≤ ax cos B + bx cos C + cx cos A

and

ax cos A + bx cos B + cx cos C ≤ ax cos C + bx cos A + cx cos B.

Hence3(ax cos A + bx cos B + cx cos C)

≤ (ax + bx + cx)(cos A + cos B + cos C).

3. Let x, y, z be positive real numbers.

(a) Prove that

x√1 + x2

+ y√1 + y2

+ z√1 + z2

≤ 3√

3

2

if x + y + z = xyz;

(b) Prove thatx

1 − x2 + y

1 − y2 + z

1 − z2 ≥ 3√

3

2

if 0 < x, y, z < 1 and xy + yz + zx = 1.

Solution: Both problems can be solved by trigonometric substitutions.

(a) By Introductory Problem 20(a), there is an acute triangle ABC withtan A = x, tan B = y, and tan C = z. Note that

tan A√1 + tan2 A

= tan A

sec A= sin A.

The desired inequality becomes

sin A + sin B + sin C ≤ 3√

3

2,

which is Introductory Problem 28(c).


(b) From the given condition and Introductory Problem 19(a), we can assumethat there is an acute triangle ABC such that

tanA

2= x, tan

B

2= y, tan

C

2= z.

By the double-angle formulas, it suffices to prove that

tan A + tan B + tan C ≥ 3√

3,

which is Introductory Problem 20(b).

4. [China 1997] Let x, y, z be real numbers with x ≥ y ≥ z ≥ π12 such that

x + y + z = π2 . Find the maximum and the minimum values of the product

cos x sin y cos z.

Solution: Let p = cos x sin y cos z. Because π2 ≥ y ≥ z, sin(y − z) ≥ 0. By

the product-to-sum formulas, we have

p = 1

2cos x[sin(y + z) + sin(y − z)] ≥ 1

2cos x sin(y + z) = 1

2cos2 x.

Note that x = π2 − (y + z) ≤ π

2 − 2 · π12 = π

3 . Hence the minimum value ofp is 1

2 cos2 π3 = 1

8 , obtained when x = π3 and y = z = π

12 .

On the other hand, we also have

p = 1

2cos z[sin(x + y) − sin(x − y)] ≤ 1

2cos2 z,

by noting that sin(x − y) ≥ 0 and sin(x + y) = cos z. By the double-angleformulas, we deduce that

p ≤ 1

4(1 + cos 2z) ≤ 1

4

(1 + cos

π

6

)= 2 + √

3

8.

This maximum value is obtained if and only if x = y = 5π24 and z = π

12 .

5. Let ABC be an acute-angled triangle, and for n = 1, 2, 3, let

xn = 2n−3(cosn A + cosn B + cosn C) + cos A cos B cos C.

Prove that

x1 + x2 + x3 ≥ 3

2.


Solution: By the arithmetic–geometric means inequality,

cos3 x + cos x

4≥ cos2 x

for x such that cos x ≥ 0. Because triangle ABC is acute, cos A, cos B, andcos C are nonnegative. Setting x = A, x = B, x = C and adding the resultinginequalities yields

x1 + x3 ≥ cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 2x2.

Consequently,

x1 + x2 + x3 ≥ 3x2 = 3

2,

by Introductory Problem 24(d).

6. Find the sum of all x in the interval [0, 2π ] such that

3 cot2 x + 8 cot x + 3 = 0.

Solution: Consider the quadratic equation

3u2 + 8u + 3 = 0.

The roots of the above equation are u1 = −8+2√

76 and u2 = −8−2

√7

6 . Bothroots are real, and their product u1u2 is equal to −1 (by Viète’s theorem).

Because y = cot x is a bijection from the interval (0, π) to the real numbers,there is a unique pair of numbers x1,1 and x2,1 with 0 < x1,1, x2,1 < π

such that cot x1,1 = u1 and cot x2,1 = u2. Because u1, u2 are negative, π2 <

x1,1, x2,1 < π , and so π < x1,1 + x2,1 < 2π . Because cot x tan x = 1 andboth tan x and cot x have period π , it follows that

1 = cot x tan x = cot x cot(π

2− x)

= cot x cot

(3π

2− x

)= cot x1,1 cot x2,1.

Therefore, x1,1+x2,1 = 3π2 . Likewise, in the interval (π, 2π), there is a unique

pair of numbers x1,2 and x2,2 satisfying the conditions of the problem withx1,2 +x2,2 = 7π

2 . Thus the answer to the problem is x1,1 +x2,1 +x1,2 +x2,2 =5π .


7. Let ABC be an acute-angled triangle with area K . Prove that√a2b2 − 4K2 +

√b2c2 − 4K2 +

√c2a2 − 4K2 = a2 + b2 + c2

2.

Solution: We have 2K = ab sin C = bc sin A = ca sin B. The expression onthe left-hand side of the desired equation is equal to√

a2b2 − a2b2 sin2 C +√

b2c2 − b2c2 sin2 A +√

c2a2 − c2a2 sin2 B

= ab cos C + bc cos A + ca cos B

= a

2(b cos C + c cos B) + b

2(c cos A + a cos C)

+ c

2(a cos B + b cos A)

= a

2· a + b

2· b + c

2· c,

and the conclusion follows.

Note: We encourage the reader to explain why this problem is the equalitycase of Advanced Problem 42(a).

8. Compute the sums(n

1

)sin a +

(n

2

)sin 2a + · · · +

(n

n

)sin na

and (n

1

)cos a +

(n

2

)cos 2a + · · · +

(n

n

)cos na.

Solution: Let Sn and Tn denote the first and second sums, respectively. Setthe complex number z = cos a + i sin a. Then, by de Moivre’s formula, wehave zn = cos na + i sin na. By the binomial theorem, we obtain

1 + Tn + iSn = 1 +(

n

1

)(cos a + i sin a) +

(n

2

)(cos 2a + i sin 2a)

+ · · · +(

n

n

)(cos na + i sin na)

=(

n

0

)z0 +

(n

1

)z +(

n

2

)z2 + · · · +

(n

n

)zn

= (1 + z)n.


Because

1 + z = 1 + cos a + i sin a = 2 cos2 a

2+ 2i sin

a

2cos

a

2

= 2 cosa

2

(cos

a

2+ i sin

a

2

),

it follows that

(1 + z)n = 2n cosn a

2

(cos

na

2+ i sin

na

2

),

again by de Moivre’s formula. Therefore,

(1 + Tn) + iSn =(

2n cosn a

2cos

na

2

)+ i(

2n cosn a

2sin

na

2

),

and so

Sn = 2n cosn a

2sin

na

2and Tn = −1 + 2n cosn a

2cos

na

2.

9. [Putnam 2003] Find the minimum value of

| sin x + cos x + tan x + cot x + sec x + csc x|

for real numbers x.

Solution: Set a = sin x and b = cos x. We want to minimize

P =∣∣∣∣a + b + a

b+ b

a+ 1

a+ 1

b

∣∣∣∣=∣∣∣∣ab(a + b) + a2 + b2 + a + b

ab

∣∣∣∣ .Note that a2 +b2 = sin2 x +cos2 x = 1. Set c = a+b. Then c2 = (a+b)2 =1+2ab, and so 2ab = c2 −1. Note also that by the addition and subtractionformulas, we have

c = sin x + cos x = √2

(√2

2sin x +

√2

2cos x

)= √

2 sin(π

4+ x)

,


and so the range of c is the interval [−√2,

√2]. Consequently, it suffices to

find the minimum of

P(c) =∣∣∣∣2ab(a + b) + 2 + 2(a + b)

2ab

∣∣∣∣=∣∣∣∣c(c2 − 1) + 2(c + 1)

c2 − 1

∣∣∣∣ = ∣∣∣∣c + 2

c − 1

∣∣∣∣=∣∣∣∣c − 1 + 2

c − 1+ 1

∣∣∣∣ .for c in the interval [−√

2,√

2]. If c−1 > 0, then by the arithmetic–geometricmeans inequality, (c−1)+ 2

c−1 > 2√

2, and so P(c) > 1+2√

2. If c−1 < 0,then by the same token,

(c − 1) + 2

c − 1= −(

(1 − c) + 2

1 − c

)≤ −2

√2,

with equality if and only if 1 − c = 21−c

, or c = 1 − √2. It follows that the

minimum value sought is∣∣∣−2

√2 + 1

∣∣∣ = 2√

2−1, obtained when c = 1−√2.

Note: Taking the derivative of the function

f (x) = sin x + cos x + tan x + cot x + sec x + csc x

and considering only its critical points is a troublesome approach to this prob-lem, because it is difficult to show that f (x) does not cross the x axis smoothly.Indeed, with a little bit more work, we can show that f (x) �= 0 with the pre-sented solution.

10. [Belarus 1999] Two real sequences x1, x2, . . . and y1, y2, . . . are defined inthe following way:

x1 = y1 = √3, xn+1 = xn +

√1 + x2

n, yn+1 = yn

1 +√1 + y2n

,

for all n ≥ 1. Prove that 2 < xnyn < 3 for all n > 1.

Solution: Writing xn = tan an for 0◦ < an < 90◦, by the half-angle formulawe have

xn+1 = tan an +√

1 + tan2 an = tan an + sec an

= 1 + sin an

cos an

= tan

(90◦ + an

2

).


Because a1 = 60◦, we have a2 = 75◦, a3 = 82.5◦, and in general an =90◦ − 30◦

2n−1 . Thus

xn = tan

(90◦ − 30◦

2n−1

)= cot

(30◦

2n−1

)= cot θn,

where θn = 30◦2n−1 .

A similar calculation shows that

yn = tan 2θn = 2 tan θn

1 − tan2 θn

,

implying that

xnyn = 2

1 − tan2 θn

.

Because 0◦ < θn < 45◦, we have 0 < tan2 θn < 1 and xnyn > 2. For n > 1,

we have θn < 30◦, implying that tan2 θn < 13 and xnyn < 3.

11. Let a, b, c be real numbers such that

sin a + sin b + sin c ≥ 3

2.

Prove that

sin(a − π

6

)+ sin

(b − π

6

)+ sin

(c − π

6

)≥ 0.

Solution: Assume for contradiction that

sin(a − π

6

)+ sin

(b − π

6

)+ sin

(c − π

6

)< 0.

Then by the addition and subtraction formulas, we have

1

2(cos a + cos b + cos c) >

√3

2(sin a + sin b + sin c) ≥ 3

√3

4.

It follows that

cos a + cos b + cos c >3√

3

2,


which implies that

sin(a + π

3

)+ sin

(b + π

3

)+ sin

(c + π

3

)= 1

2(sin a + sin b + sin c) +

√3

2(cos a + cos b + cos c)

>1

2· 3

2+

√3

2· 3

√3

2= 3,

which is impossible, because sin x < 1.

12. Consider any four numbers in the interval[√

2−√6

2 ,√

2+√6

2

]. Prove that there

are two of them, say a and b, such that∣∣∣a√4 − b2 − b√

4 − a2∣∣∣ ≤ 2.

Solution: Dividing both sides of the inequality by 4 yields∣∣∣∣∣∣a2√

1 −(

b

2

)2

− b

2

√1 −(a

2

)2

∣∣∣∣∣∣ ≤ 1

2.

We substitute a2 = sin x and b

2 = sin y. The last inequality reduces to

| sin(x − y)| = | sin x cos y − sin y cos x| ≤ sinπ

6. (∗)

We want to find t1 and t2 such that

sin t1 =√

2 − √6

4and sin t2 =

√2 + √

6

4.

By the double-angle formulas, we conclude that cos 2t1 = 1 − 2 sin2 t1

= 1 − 8−4√

38 =

√3

2 = cos(±π

6

)and cos 2t2 = −

√3

2 = cos 5π6 . Because

y = sin x is a one-to-one and onto map between the intervals[−π

2 , π2

]and

[−1, 1], it follows that t1 = − π12 and t2 = 5π

12 .

We divide the interval[− π

12 , 5π12

]into three disjoint intervals of length π

6 :

I1 = [− π12 , π

12

), I2 = [ π

12 , π4

), and I3 =

[π4 , 5π

12

]. The function y = 2 sin x

takes the intervals I1, I2, I3 injectively and surjectively to the intervals I ′1 =

5. Solutions to Advanced Problems 135[√2−√

62 , 2 sin π

12

), I2 =

[2 sin π

12 ,√

2)

, I ′3 =[√

2,√

2+√6

2

], respectively.

By the pigeonhole principle, one of the intervals I ′1, I

′2, or I ′

3 contains two ofthe four given numbers, say a and b. It follows that one of the intervals I1, I2,or I3 contains x and y such that a = 2 sin x and b = 2 sin y. Because theintervals I1, I2, and I3 have equal lengths of π

6 , it follows that |x − y| ≤ π6 .

We have obtained the desired the inequality (∗).

13. Let a and b be real numbers in the interval [0, π2 ]. Prove that

sin6 a + 3 sin2 a cos2 b + cos6 b = 1

if and only if a = b.

Solution: The first equality can be rewritten as

(sin2 a)3 + (cos2 b)3 + (−1)3 − 3(sin2 a)(cos2 b)(−1) = 0. (∗)

We will use the identity

x3 + y3 + z3 − 3xyz = 1

2(x + y + z)[(x − y)2 + (y − z)2 + (z − x)2].

Let x = sin2 a, y = cos2 b, and z = −1. According to equation (∗) we havex3+y3+z3−3xyz = 0. Hence x+y+z = 0 or (x−y)2+(y−z)2+(z−x)2 =0. The latter would imply x = y = z, or sin2 a = cos2 b = −1, whichis impossible. Thus x + y + z = 0, so that sin2 a + cos2 b − 1 = 0, orsin2 a = 1 − cos2 b. It follows that sin2 a = sin2 b, and taking into accountthat 0 ≤ a, b ≤ π

2 , we obtain a = b.

Even though all the steps above are reversible, we will show explicitly that ifa = b, then

sin6 a + cos6 a + 3 sin2 a cos2 b = 1.

Indeed, the expression on the left-hand side could be written as

(sin2 a + cos2 a)(sin4 a − sin2 a cos2 a + cos4 a) + 3 sin2 a cos2 a

= (sin2 a + cos2 a)2 − 3 sin2 a cos2 a + 3 sin2 a cos2 a = 1.

14. Let x, y, z be real numbers with 0 < x < y < z < π2 . Prove that

π

2+ 2 sin x cos y + 2 sin y cos z ≥ sin 2x + sin 2y + sin 2z.


Solution: By the double-angle formulas, the above inequalities reduce to

π

2> 2 sin x(cos x − cos y) + 2 sin y(cos y − cos z) + 2 sin z cos z,

orπ

4> sin x(cos x − cos y) + sin y(cos y − cos z) + sin z cos z.

As shown in Figure 5.1, in the rectangular coordinate plane, we consider pointsO = (0, 0), A = (cos x, sin x), A1 = (cos x, 0), B = (cos y, sin y), B1 =(cos y, 0), B2 = (cos y, sin x), C = (cos z, sin z), C1 = (cos z, 0), C2 =(cos z, sin y), and D = (0, sin z). Points A, B, and C are in the first quadrantof the coordinate plane, and they lie on the unit circle in counterclockwiseorder.

A

B

CD

O A1B1C1

B2

C2

Figure 5.1.

Let D denote the region enclosed by the unit circle in the first quadrant (in-cluding the boundary). It is not difficult to see that quadrilaterals AA1B1B2,BB1C1C2, and CC1OD are nonoverlapping rectangles inside region D. It isalso not difficult to see that [D] = π

4 , [AA1B1B2] = sin x(cos x − cos y),[BB1C1C2] = sin y(cos y − cos z), and [CC1OD] = sin z cos z, from whichour desired result follows.

15. For a triangle XYZ, let rXYZ denote its inradius. Given that the convex pen-tagon ABCDE is inscribed in a circle, prove that if rABC = rAED andrABD = rAEC , then triangles ABC and AED are congruent.

Solution: Let R be the radius of the circle in which ABCDE is inscribed.


A

B

C

D

E

2a

2b

2c

2d

2e

Figure 5.2.

As shown in the proof of Problem 27(a), if ABC is a triangle with inradius r

and circumradius R, then

1 + r

R= cos A + cos B + cos C = cos A − cos(A + C) + cos C.

Let 2a, 2b, 2c, 2d , and 2e be the measures of arcs AB, BC, CD, DE, andEA, respectively. Then a + b + c + d + e = 180◦. Because rABC = rAED

and rABD = rAEC , we have

cos a − cos(a + b) + cos b = cos d + cos e − cos(d + e) (∗)

and

cos a + cos(b + c) + cos(d + e) = cos e + cos(c + d) + cos(a + b).

Subtracting the two equations, we obtain cos b+cos(c+d) = cos d +cos(b+c), or

2 cosb + c + d

2cos

b − c − d

2= 2 cos

b + c + d

2cos

d − b − c

2

by sum-to-product formulas. It follows that

cosb − c − d

2= cos

d − b − c

2,

and so b = d. Plugging this result into equation (∗) yields

cos a − cos(a + b) + cos b = cos b + cos e − cos(b + e),


or cos a + cos(b + e) = cos e + cos(a + b). Applying the sum-to-productformulas again gives

2 cosa + b + e

2cos

a − b − e

2= 2 cos

a + b + e

2cos

e − a − b

2,

and so cos a−b−e2 = cos e−a−b

2 . It follows that a = e. Because a = e andb = d, triangles ABC and AED are congruent.

16. All the angles in triangle ABC are less then 120◦. Prove that

cos A + cos B − cos C

sin A + sin B − sin C> −

√3

3.

Solution: Consider the triangle A1B1C1, as shown in Figure 5.3, where � A1 =120◦ − � A, � B1 = 120◦ − � B, and � C1 = 120◦ − � C. The given conditionguarantees the existence of such a triangle.

A B

C A1

B1

C1

Figure 5.3.

Applying the triangle inequality in triangle A1B1C1 gives B1C1 + C1A1 >

A1B1; that issin A1 + sin B1 > sin C1

by applying the law of sines to triangle A1B1C1. It follows that

sin(120◦ − A) + sin(120◦ − B) > sin(120◦ − C),

or √3

2(cos A + cos B − cos C) + 1

2(sin A + sin B − sin C) > 0.

Taking into account that a + b > c implies sin A + sin B − sin C > 0, theabove inequality can be rewritten as

√3

2· cos A + cos B − cos C

sin A + sin B − sin C+ 1

2> 0,

from which the conclusion follows.


17. [USAMO 2002] Let ABC be a triangle such that(cot

A

2

)2

+(

2 cotB

2

)2

+(

3 cotC

2

)2

=(

6s

7r

)2

,

where s and r denote its semiperimeter and its inradius, respectively. Provethat triangle ABC is similar to a triangle T whose side lengths are all positiveintegers with no common divisor and determine these integers.

Solution: Define

u = cotA

2, v = cot

B

2, w = cot

C

2.

As shown in Figure 5.4, denote the incenter of triangle ABC by I , and letD, E, and F be the points of tangency of the incircle with sides BC, CA, andAB, respectively. Then |EI | = r , and by the standard formula, |AE| = s −a.

A

B CD

E

F

I

Figure 5.4.

We have

u = cotA

2= |AE|

|EI | = s − a

r,

and similarly v = s−br

, w = s−cr

. Because

s

r= (s − a) + (s − b) + (s − c)

r= u + v + w,

we can rewrite the given relation as

49[u2 + 4v2 + 9w2

]= 36(u + v + w)2.


Expanding the last equality and canceling like terms, we obtain

13u2 + 160v2 + 405w2 − 72(uv + vw + wu) = 0,

or(3u − 12v)2 + (4v − 9w)2 + (18w − 2u)2 = 0.

Therefore, u : v : w = 1 : 14 : 1

9 . This can also be realized by recognizing thatthe given relation corresponds to equality in Cauchy–Schwarz inequality

(62 + 32 + 22)[u2 + (2v)2 + (3w)2

]≥ (6 · u + 3 · 2v + 2 · 3w)2.

After multiplying by r , we see that

s − a

36= s − b

9= s − c

4= 2s − b − c

9 + 4= 2s − c − a

4 + 36= 2s − a − b

36 + 9

= a

13= b

40= c

45;

that is, triangle ABC is similar to a triangle with side lengths 13, 40, 45.

Note: The technique of using

a

b= c

d= a + c

b + d

is rather tricky. However, by Introductory Problem 19(a), we can have

u + v + w = uvw.

Since u : v : w = 1 : 14 : 1

9 , it follows that u = 7, v = 74 , and w = 7

9 . Henceby the double-angle formulas, sin A = 7

25 , sin B = 5665 , and sin C = 63

65 , or

sin A = 133257

, sin B = 403257

, sin C = 453257

.

By the extended law of sines, triangle ABC is similar to triangle T with sidelengths 13, 40, and 45. (The circumcircle of T has diameter 325

7 .)

18. [USAMO 1996] Prove that the average of the numbers

2 sin 2◦, 4 sin 4◦, 6 sin 6◦, . . . , 180 sin 180◦,

is cot 1◦.


First Solution: We need to prove that

2 sin 2◦ + 4 sin 4◦ + · · · + 178 sin 178◦ = 90 cot 1◦,

which is equivalent to

2 sin 2◦ · sin 1◦ + 2(2 sin 4◦ · sin 1◦) + · · · + 89(2 sin 178◦ · sin 1◦)= 90 cos 1◦.

Note that2 sin 2k◦ sin 1◦ = cos(2k − 1)◦ − cos(2k + 1)◦.

We have

2 sin 2◦ · sin 1◦ + 2(2 sin 4◦ · sin 1◦) + · · · + 89(2 sin 178◦ · sin 1◦)= (cos 1◦ − cos 3◦) + 2(cos 3◦ − cos 5◦)

+ · · · + 89(cos 177◦ − cos 179◦)= cos 1◦ + cos 3◦ + · · · + cos 177◦ − 89 cos 179◦

= cos 1◦ + (cos 3◦ + cos 177◦) + · · · + (cos 89◦ + cos 91◦)+ 89 cos 1◦

= cos 1◦ + 89 cos 1◦ = 90 cos 1◦,

as desired.

Note: The techniques of telescoping sum and pairing of summands involvedin the first solution is rather tricky. The second solution involves complexnumbers. It is slightly longer than the first solution. But for the reader who isfamiliar with the rules of operation for complex numbers and geometricseries, every step is natural.

Second Solution: Set the complex number z = cos 2◦ + i sin 2◦. Then, byde Moivre’s formula, we have zn = cos 2n◦ + i sin 2n◦. Let a and b be realnumbers such that

z + 2z2 + · · · + 89z89 = a + bi.

Because sin 180◦ = 0,

b = 1

2(2 sin 2◦ + 4 sin 4◦ + · · · + 178 sin 178◦ + 180 sin 180◦),

and it suffices to show that b = 45 cot 1◦.


Setpn(x) = x + 2x2 + · · · + nxn.

Then

(1 − x)pn(x) = pn(x) − xpn(x) = x + x2 + · · · + xn − nxn+1.

Setqn(x) = (1 − x)pn(x) + nxn+1 = x + x2 + · · · + xn.

Then (1 − x)qn(x) = qn(x) − xqn(x) = x − xn+1. Consequently, we have

pn(x) = qn(x)

1 − x− nxn+1

1 − x= x − xn+1

(1 − x)2 − nxn+1

1 − x.

It follows that

a + bi = z + 2z2 + · · · + 89z89 = p89(z)

= z − z90

(1 − z)2 − 89z90

1 − z= z + 1

(z − 1)2 − 89

z − 1,

because z90 = cos 180◦+i sin 180◦ = −1. Note that z+1 = cis 2◦+cis 0◦ =2 cos 1◦ cis 1◦ and z − 1 = cis 2◦ − cis 0◦ = 2 sin 1◦ cis 91◦, and so

a + bi = 2 cos 1◦ cis 1◦

(2 sin 1◦ cis 91◦)2 − 89

2 sin 1◦ cis 91◦

= 2 cos 1◦ cis 1◦

4 sin2 1◦ cis 182◦ − 89 cis(−91◦)2 sin 1◦

= cos 1◦ cis(−181◦)2 sin2 1◦ − 89 cis(−91◦)

2 sin 1◦ .

Therefore,

b = cos 1◦ sin(−181◦)2 sin2 1◦ − 89 sin(−91◦)

2 sin 1◦ = cos 1◦ sin 1◦

2 sin2 1◦ + 89 cos 1◦

2 sin 1◦

= cos 1◦

2 sin 1◦ + 89 cos 1◦

2 sin 1◦ = 45 cot 1◦,

as desired.

19. Prove that in any acute triangle ABC,

cot3 A + cot3 B + cot3 C + 6 cot A cot B cot C ≥ cot A + cot B + cot C.


Solution: Let cot A = x, cot B = y, and cot C = z. Because xy+yz+zx = 1(Introductory Problem 21), it suffices to prove the homogeneous inequality

x3 + y3 + z3 + 6xyz ≥ (x + y + z)(xy + yz + zx).

But this is equivalent to

x(x − y)(x − z) + y(y − z)(y − x) + z(z − x)(z − y) ≥ 0,

which is Schur’s inequality.

20. [Turkey 1998] Let {an} be the sequence of real numbers defined by a1 = t

and an+1 = 4an(1 − an) for n ≥ 1. For how many distinct values of t do wehave a1998 = 0?

Solution: Let f (x) = 4x(1−x) = 1−(2x−1)2. Observe that if 0 ≤ f (x) ≤1, then 0 ≤ x ≤ 1. Hence if a1998 = 0, then we must have 0 ≤ t ≤ 1. Nowchoose 0 ≤ θ ≤ π

2 such that sin θ = √t . Observe that for any φ ∈ R,

f (sin2 φ) = 4 sin2 φ (1 − sin2 φ) = 4 sin2 φ cos2 φ = sin2 2φ;since a1 = sin2 θ , it follows that

a2 = sin2 2θ, a3 = sin2 4θ, . . . , a1998 = sin2 21997θ.

Therefore, a1998 = 0 if and only if sin 21997θ = 0. That is, θ = kπ21997 for some

integers k, and so the values of t for which a1998 = 0 are sin2(kπ/21997), wherek ∈ Z. Therefore we get 21996 + 1 such values of t , namely, sin2(kπ/21997)

for k = 0, 1, 2, · · · , 21996.

21. Triangle ABC has the following property: there is an interior point P suchthat � PAB = 10◦, � PBA = 20◦, � PCA = 30◦, and � PAC = 40◦. Provethat triangle ABC is isosceles.

Solution: Consider Figure 5.5, in which all angles are in degrees.

A

B

C

P

Figure 5.5.


Let x = � PCB (in degrees). Then � PBC = 80◦ − x. By the law of sines orby Ceva’s theorem,

1 = PA

PB· PB

PC· PC

PA= sin � PBA

sin � PAB· sin � PCB

sin � PBC· sin � PAC

sin � PCA

= sin 20◦ sin x sin 40◦

sin 10◦ sin(80◦ − x) sin 30◦ = 4 sin x sin 40◦ cos 10◦

sin(80◦ − x).

The product-to-sum formulas yield

1 = 2 sin x(sin 30◦ + sin 50◦)sin(80◦ − x)

= sin x(1 + 2 cos 40◦)sin(80◦ − x)

,

and so

2 sin x cos 40◦ = sin(80◦ − x) − sin x = 2 sin(40◦ − x) cos 40◦,

by the difference-to-product formulas. We conclude that x = 40◦ − x, orx = 20◦. It follows that � ACB = 50◦ = � BAC, and so triangle ABC isisosceles.

22. Let a0 = √2 + √

3 + √6, and let an+1 = a2

n−52(an+2)

for integers n > 0. Provethat

an = cot

(2n−3π

3

)− 2

for all n.

Solution: By either the double-angle or the half-angle formulas, we obtain

cotπ

24= cos π

24

sin π24

= 2 cos2 π24

2 sin π24 cos π

24

= 1 + cos π12

sin π12

= 1 + cos(

π3 − π

4

)sin(

π3 − π

4

) = 1 + cos π3 cos π

4 + sin π3 sin π

4

sin π3 cos π

4 − cos π3 sin π

4

= 1 +√

24 +

√6

4√6

4 −√

24

= 4 + √6 + √

2√6 − √

2

= 4(√

6 + √2) + (

√6 + √

2)2

(√

6 − √2)(

√6 + √

2)= 4(

√6 + √

2) + 8 + 4√

3

4

= 2 + √2 + √

3 + √6 = a0 + 2.

Hence an = cot(

2n−3π3

)− 2 is true for n = 0.


It suffices to show that bn = cot(

2n−3π3

), where bn = an + 2, n ≥ 1. The

recursive relation becomes

bn+1 − 2 = (bn − 2)2 − 5

2bn

,

or

bn+1 = b2n − 1

2bn

.

Assuming, inductively, that bk = cot ck , where ck = 2k−3π3 , yields

bk+1 = cot2 ck − 1

2 cot ck

= cot 2ck = cot ck+1,

and we are done.

23. [APMC 1982] Let n be an integer with n ≥ 2. Prove that

n∏k=1

tan

[π

3

(1 + 3k

3n − 1

)]=

n∏k=1

cot

[π

3

(1 − 3k

3n − 1

)].

Solution: Let

uk = tan

[π

3

(1 + 3k

3n − 1

)]and vk = tan

[π

3

(1 − 3k

3n − 1

)].

The desired equality becomes

n∏k=1

ukvk = 1. (∗)

Set

tk = tan3k−1π

3n − 1.

Applying the addition and and subtraction formulas yields

uk = tan

(π

3+ 3k−1π

3n − 1

)=

√3 + tk

1 − √3tk

and vk =√

3 − tk

1 + √3tk

.

The triple-angle formulas give

tk+1 = 3tk − t3k

1 − 3t2k

,


implying that

tk+1

tk= 3 − t2

k

1 − 3t2k

=√

3 + tk

1 − √3tk

·√

3 − tk

1 + √3tk

= ukvk.

Consequently,

n∏k=1

(ukvk) = t2

t1· t3

t2· · · tn+1

tn=

tan(π + π

3n−1

)tan(

π3n−1

) = 1,

establishing equation (∗).

24. [China 1999, by Yuming Huang] Let P2(x) = x2 − 2. Find all sequences ofpolynomials {Pk(x)}∞k=1 such that Pk(x) is monic (that is, with leading coef-ficient 1), has degree k, and Pi(Pj (x)) = Pj (Pi(x)) for all positive integers i

and j .

Solution: First, we show that the sequence, if it exists, is unique. In fact, foreach n, there can be only one Pn that satisfies Pn(P2(x)) = P2(Pn(x)). Let

Pn(x) = xn + an−1xn−1 + · · · + a1x + a0.

By assumption,

(x2 − 2)n + an−1(x2 − 2)n−1 + · · · + a1(x

2 − 2) + a0

= (xn + an−1xn−1 + · · · + a1x + a0)

2 − 2.

Consider the coefficients on both sides. On the left side, 2i is the highest powerof x in which ai appears. On the right side, the highest power is xn+i , and thereit appears as 2aix

n+i . Thus, we see that the maximal power of ai always ishigher on the right side. It follows that we can solve for each ai in turn, fromn − 1 to 0, by equating coefficients. Furthermore, we are guaranteed that thepolynomial is unique, since the equation we need to solve to find each ai islinear.

Second, we define Pn explicitly. We claim that Pn(x) = 2Tn

(x2

)(where Tn is

the nth Chebyshev polynomial defined in Introductory Problem 49). That is,Pn is defined by the recursive relation P1(x) = x, P2(x) = x2 − 2, and

Pn+1(x) = xPn(x) − Pn−1(x).


Because Tn(cos θ) = cos nθ , Pn(2 cos θ) = 2 cos nθ . It follows that for all θ ,

Pm(Pn(2 cos θ)) = Pm(2 cos nθ) = 2 cos mnθ

= Pn(2 cos mθ) = Pn(Pm(2 cos θ)).

Thus, Pm(Pn(x)) and Pn(Pm(x)) agree at all values of x in the interval [−2, 2].Because both are polynomials, it follows that they are equal for all x, whichcompletes the proof.

25. [China 2000, by Xuanguo Huang] In triangle ABC, a ≤ b ≤ c. As a functionof angle C, determine the conditions under which a +b−2R −2r is positive,negative, or zero.

Solution: In Figure 5.6, set � A = 2x, � B = 2y, � C = 2z. Then 0 < x ≤y ≤ z and x + y + z = π

2 . Let s denote the given quantity a + b − 2R − 2r .Using the extended law of sines and by Introductory Problem 25(d), we have

s = 2R(sin 2x + sin 2y − 1 − 4 sin x sin y sin z).

A

B

C

Figure 5.6.

Note that in a right triangle ABC with � C = π2 , we have 2R = c and

2r = a + b − c, implying that s = 0. Hence, we factor cos 2z from ourexpression for s. By the sum-to-product, product-to-sum, and double-angle


formulas, we haves

2R= 2 sin (x + y) cos (x − y) − 1 + 2(cos (x + y) − cos (x − y)) sin z

= 2 cos z cos (x − y) − 1 + 2(sin z − cos (x − y)) sin z

= 2 cos (x − y)(cos z − sin z) − cos 2z

= 2 cos (y − x) · cos2 z − sin2 z

cos z + sin z− cos 2z

=[

2 cos (y − x)

cos z + sin z− 1

]cos 2z,

where we may safely introduce the quantity cos z+ sin z because it is positivewhen 0 < z < π

2 .

Observe that 0 ≤ y − x < min{ y, x + y } ≤ min{z, π

2 − z}. Because

z ≤ π2 and π

2 − z ≤ π2 , we have cos (y − x) > max

{cos z, cos

(π2 − z

) } =max{ cos z, sin z }. Hence 2 cos(y − x) > cos z + sin z, or

2 cos (x − y)

cos z + sin z− 1 > 0.

Thus, s = p cos 2z for some p > 0. It follows that s = a + b − 2R − 2r

is positive, zero, or negative if and only if angle C is acute, right, or obtuse,respectively.

26. Let ABC be a triangle. Points D, E, F are on sides BC, CA, AB, respectively,such that |DC| + |CE| = |EA| + |AF | = |FB| + |BD|. Prove that

|DE| + |EF | + |FD| ≥ 1

2(|AB| + |BC| + |CA|).

Solution: As shown in Figure 5.7, Let E1 and F1 be the feet of the perpen-dicular line segments from E and F to line BC.

A

B CD

EF

B CD

E

F

A

E1 E1F1 F1

Figure 5.7.


We have

|EF | ≥ |E1F1| = a − (|BF | cos B + |CE| cos C).

Likewise, we have

|DE| ≥ c − (|AE| cos A + |BD| cos B)

and|FD| ≥ b − (|CD| cos C + |AF | cos A).

Note that |DC| + |CE| = |EA| + |AF | = |FB| + |BD| = 13 (a + b + c).

Adding the last three inequalities gives

|DE| + |EF | + |FD|≥ a + b + c − 1

3(a + b + c)(cos A + cos B + cos C)

≥ 1

2(a + b + c),

by Introductory Problem 27(b). Equality holds if and only if the length ofsegment EF (FD and DE) is equal to the length of the projection of segmentEF on line BC (FD on CA and DE on AB), and A = B = C = 60◦, thatis, if and only if D, E, and F are the midpoints of an equilateral triangle.

27. Let a and b be positive real numbers. Prove that

1√1 + a2

+ 1√1 + b2

≥ 2√1 + ab

if either (1) 0 < a, b ≤ 1 or (2) ab ≥ 3.

Note: Part (1) appeared in the Russian Mathematics Olympiad in 2001.

Solution: Because a and b are positive real numbers, there are angles x andy, with 0 < x, y < 90◦, such that tan x = a and tan y = b. The desiredinequality is clearly true when a = b. Hence we assume that a �= b, orequivalently, x �= y. Then 1 + a2 = sec2 x and 1√

1+a2 = cos x. Note that

1 + ab = cos x cos y + sin x sin y

cos x cos y= cos(x − y)

cos x cos y

by the addition and subtraction formulas. The desired inequality reduces to

cos x + cos y ≥ 2√

cos x cos y

cos(x − y). (∗)


To establish part (1), we rewrite inequality (∗) as

cos2 x + cos2 y + 2 cos x cos y ≤ 4 cos x cos y

cos(x − y).

Because 0 < |x − y| < 90◦, it follows that 0 < cos(x − y) < 1. Hence2 cos x cos y ≤ 2 cos x cos y

cos(x−y). It suffices to show that

cos(x − y)[cos2 x + cos2 y

]≤ 2 cos x cos y,

orcos(x − y) [cos 2x + cos 2y + 2] ≤ 4 cos x cos y

by the double-angle formulas. By the sum-to-product formulas, the lastinequality is equivalent to

cos(x − y)[2 cos(x − y) cos(x + y) + 2] ≤ 2[cos(x − y) + cos(x + y)],or cos2(x − y) cos(x + y) ≤ cos(x + y), which is clearly true, because for0 < a, b ≤ 1, we have 0◦ < x, y < 45◦, and so 0◦ < x + y ≤ 90◦ andcos(x + y) > 0. This completes the proof of part (1).

To prove part (2), we rewrite inequality (∗) as

2 cosx + y

2cos

x − y

2≥ 2

√12 [cos(x + y) + cos(x − y)]

cos(x − y)

by the sum-to-product and product-to-sum formulas. Squaring both sides ofthe inequality and clearing denominators gives

4 cos2 x + y

2cos2 x − y

2cos(x − y) ≥ 2[cos(x + y) + cos(x − y)],

or

[1 + cos(x + y)][1 + cos(x − y)] cos(x − y) ≥ 2[cos(x + y) + cos(x − y)]by the double-angle formulas. Setting s = cos(x + y) and t = cos(x − y), itsuffices to prove that

(1 + s)(1 + t)t ≥ 2(s + t),

or,0 ≤ (1 + s)t2 + (s − 1)t − 2s = (t − 1)[(1 + s)t + 2s].

Because t ≤ 1, it suffices to show that

(1 + s)t + 2s ≤ 0.


Because ab ≥ 3, tan x tan y ≥ 3, or equivalently, sin x sin y ≥ 3 cos x cos y.By the product-to-sum formulas, we have

1

2[cos(x − y) − cos(x + y)] ≥ 3

2[cos(x − y) + cos(x + y)],

or t ≤ −2s. Because 1 + s ≥ 0, (1 + s)t ≤ −(1 + s)2s. Consequently,(1 + s)t + 2s ≤ −(1 + s)2s + 2s = −2s2 ≤ 0, as desired.

28. [China 1998, by Xuanguo Huang] Let ABC be a nonobtuse triangle such thatAB > AC and � B = 45◦. Let O and I denote the circumcenter and incenter oftriangle ABC, respectively. Suppose that

√2|OI | = |AB|−|AC|. Determine

all the possible values of sin A.

First Solution: Applying the extended law of sines to triangle ABC yieldsa = 2R sin A, b = 2R sin B, and c = 2R sin C. If incircle is tangent to sideAB at D (Figure 5.8). Then |BD| = c+a−b

2 , and so r = |ID| = |BD| tan B2 .

The half-angle formulas give

tanB

2= 1 − cos B

sin B= 1 −

√2

2√2

2

= √2 − 1,

and sor = R

(√2 − 1

)(sin A + sin C − sin B).

By Euler’s formula, |OI |2 = R(R − 2r), so we have

|OI |2 = R2 − 2Rr = R2[1 − 2(sin A + sin C − sin B)

(√2 − 1

)].

Squaring both sides of the given equation√

2|OI | = |AB| − |AC| gives

|OI |2 = (c − b)2

2= 2R2(sin C − sin B)2.

Therefore

2(sin C − sin B)2 = 1 − 2(sin A + sin C − sin B)(√

2 − 1)

or

1 − 2

(sin C −

√2

2

)2

= 2

(sin A + sin C −

√2

2

)(√2 − 1

). (∗)


The addition and subtraction formulas give

sin C = sin(180◦ − B − A) = sin(135◦ − A)

= sin 135◦ cos A − cos 135◦ sin A =√

2(sin A + cos A)

2,

and so

sin C −√

2

2=

√2

2(sin A + cos A − 1).

Plugging the last equation into equation (∗) yields

1 − (sin A+ cos A− 1)2 = 2(√

2 − 1)[

sin A +√

2

2(sin A + cos A − 1)

].

Expanding both sides of the last equation gives

1 − (sin A + cos A)2 + 2(sin A + cos A) − 1

=(√

2 − 1) (

2 + √2)

sin A +(

2 − √2)

(cos A − 1)

or

sin2 A+ cos2 A+ 2 sin A cos A =(

2 − √2)

sin A+√2 cos A+

(2 − √

2)

.

Consequently, we have

2 sin A cos A −(

2 − √2)

sin A − √2 cos A +

(√2 − 1

)= 0;

that is, (√2 sin A − 1

) (√2 cos A − √

2 + 1)

= 0.

This implies that sin A =√

22 or cos A = 1 −

√2

2 . Therefore, the answer to theproblem is

sin A =√

2

2or sin A =

√1 − cos2 A =

√4√

2 − 2

2.


A

B

C

A

BCD

E

D

E

F

MM

O

I

I

O (F)

Figure 5.8.

Second Solution: As shown in Figure 5.8, the incircle touches the sidesAB, BC, and CA at D, E, and F , respectively. Let M be the foot of theperpendicular line segments from O to side BC. Then the line OM is theperpendicular bisector of BC, and |BM| = |CM|. From equal tangents, wehave |AF | = |AE|, |BD| = |BF |, and |CD| = |CE|. Because c > b, M lieson segment BD. We find that

√2|OI | = c − b = (|AF | + |FB|) − (|AE| + |EC|)

= |FB| − |EC| = |BD| − |DC|.We deduce that |BD| = |BM| + |MD| and |DC| = |CM| − |DM|. Hence√

2|OI | = 2|DM|, or |OI | = √2|DM|. Thus lines OI and DM form a 45◦

angle, which implies that either OI ⊥ AB or OI ‖ AB. We consider thesetwo cases separately.

• First Case: In this case, we assume that OI ⊥ AB. Then OI is theperpendicular bisector of side AB; that is, the incenter lies on the per-pendicular bisector of side AB. Thus triangle ABC must be isosceles,

with |AC| = |BC|, and so A = B = 45◦ and sin A =√

22 .

• Second Case: In this case, we assume that OI ‖ AB. Let N be themidpoint of side AB. Then OIFN is a rectangle. Note that � AON =� C, and thus

R cos � AON = R cos C = |ON | = |IF | = r.

By the solution to Introductory Problem 27, we have

cos C = R

r= cos A + cos B + cos C − 1,


implying that cos A = 1 − cos B = 1 − √2/2. It follows that

sin A =√

1 − cos2 A =√

4√

2 − 2

2.

29. [Dorin Andrica] Let n be a positive integer. Find the real numbers a0 and ak,�,1 ≤ � < k ≤ n, such that

sin2 nx

sin2 x= a0 +

∑1≤�<k≤n

a�,k cos 2(k − �)x

for all real numbers x with x not an integer multiple of π .

Solution: In this solution, we apply a similar technique to that shown in thefirst solution of Advanced Problem 18. Note that

2 sin 2kx sin x = cos(2k − 1)x − cos(2k + 1)x.

We have

2 sin x(sin 2x + sin 4x + · · · + sin 2nx)

= [cos x − cos 3x] + [cos 3x − cos 5x]+ · · · + [cos(2n − 1)x − cos(2n + 1)x]

= cos x − cos(2n + 1)x = 2 sin nx sin(n + 1)x,

or

s = sin 2x + sin 4x + · · · + sin 2nx = sin nx sin(n + 1)x

sin x.

Similarly, by noting that

2 cos 2kx sin x = sin(2k + 1)x − sin(2k − 1)x,

we have

2 sin x(cos 2x + cos 4x + · · · + cos 2nx)

= [sin 3x − sin x] + [sin 5x − sin 3x]+ · · · + [sin(2n + 1)x − sin(2n − 1)x]

= sin(2n + 1)x − sin x = 2 sin nx cos(n + 1)x,

or

c = cos 2x + cos 4x + · · · + cos 2nx = sin nx cos(n + 1)x

sin x.


It follows that(sin2 nx

sin2 x

)2

=(

sin nx sin(n + 1)x

sin x

)2

+(

sin nx cos(n + 1)x

sin x

)2

= s2 + c2.

On the other hand,

s2 + c2 = (sin 2x + sin 4x + · · · + sin 2nx)2

+ (cos 2x + cos 4x + · · · + cos 2nx)2

= n +∑

1≤�<k≤n

(2 sin 2�x sin 2kx + 2 cos 2�x cos 2kx)

= n + 2∑

1≤�<k≤n

cos 2(k − �)x

by the product-to-sum formulas. Setting a0 = n and a�,k = 2 solves theproblem.

30. [USAMO 2000] Let S be the set of all triangles ABC for which

5

(1

|AP | + 1

|BQ| + 1

|CR|)

− 3

min{ |AP |, |BQ|, |CR| } = 6

r,

where r is the inradius and P , Q, and R are the points of tangency of theincircle with sides AB, BC, and CA, respectively. Prove that all triangles inS are isosceles and similar to one another.

Solution: Let I be the incenter of triangle ABC. Then |IP | = |IQ| =|IR| = r . By symmetry, we may assume that min{ |AP |, |BQ|, |CR| } =|AP |, as shown in Figure 5.9. Let x = tan A

2 , y = tan B2 , and z = tan C

2 . ByIntroductory Problem 19(a), we also have

xy + yz + zx = 1. (∗)

A

B C

P

Q

R

I

Figure 5.9.


Note that |AP | = rx

, |BQ| = ry

, and |CR| = rz. Then the equation given in

the problem statement becomes

2x + 5y + 5z = 6. (∗∗)

Eliminating x from equations (∗) and (∗∗) yields

5y2 + 5z2 + 8yz − 6y − 6z + 2 = 0.

Completing the squares, we obtain

(3y − 1)2 + (3z − 1)2 = 4(y − z)2.

Setting 3y − 1 = u and 3z − 1 = v gives y = u+13 and z = v+1

3 , and soy − z = u−v

3 . The above equation becomes

5u2 + 8uv + 5v2 = 0.

Because the discriminant of this quadratic equation is 82 −4 ·25 < 0, the onlyreal solution to the equation is u = v = 0. Thus there is only one possibleset of values for the tangents of half-angles of ABC (namely, x = 4

3 andy = z = 1

3 ). Thus all triangles in S are isosceles and similar to one another.

Indeed, we have x = r|AP | = 4

3 and y = z = r|BQ| = r

|CQ| = 13 = 4

12 , so wecan set r = 4, |AP | = |AR| = 3, and |BP | = |BQ| = |CQ| = |CR| = 12.This leads to |AB| = |AC| = 15 and |BC| = 24. By scaling, all triangles inS are similar to the triangle with side lengths 5, 5, 8.

We can also use the half-angle formulas to calculate

sin B = sin C = 2 tan C2

1 + tan2 C2

= 3

5.

From this it follows that |AQ| : |QB| : |BA| = 3 : 4 : 5 and |AB| : |AC| :|BC| = 5 : 5 : 8.

31. [TST 2003] Let a, b, c be real numbers in the interval (0, π2 ). Prove that

sin a sin(a − b) sin(a − c)

sin(b + c)+ sin b sin(b − c) sin(b − a)

sin(c + a)

+ sin c sin(c − a) sin(c − b)

sin(a + b)≥ 0.


Solution: By the product-to-sum and the double-angle formulas, we have

sin(α − β) sin(α + β) = 1

2[cos 2β − cos 2α]

= sin2 α − sin2 β.

Hence, we obtain

sin a sin(a − b) sin(a − c) sin(a + b) sin(a + c)

= sin a(sin2 a − sin2 b)(sin2 a − sin2 c)

and its analogous cyclic symmetric forms. Therefore, it suffices to prove that

x(x2 − y2

) (x2 − z2

)+y(y2 − z2

) (y2 − x2

)+z(z2 − x2

) (z2 − y2

)≥ 0,

where x = sin a, y = sin b, and z = sin c (hence x, y, z > 0). Since thelast inequality is symmetric with respect to x, y, and z, we may assume that0 < x ≤ y ≤ z. It suffices to prove that

x(y2 − x2

) (z2 − x2

)+z(z2 − x2

) (z2 − y2

)≥ y(z2 − y2

) (y2 − x2

),

which is evident, because

x(y2 − x2

) (z2 − x2

)≥ 0

and

z(z2 − x2

) (z2 − y2

)≥ z(y2 − x2

) (z2 − y2

)≥ y(z2 − y2

) (y2 − x2

).

Note: The key step of the proof is an instance of Schur’s inequality withr = 1

2 .

32. [TST 2002] Let ABC be a triangle. Prove that

sin3A

2+ sin

3B

2+ sin

3C

2≤ cos

A − B

2+ cos

B − C

2+ cos

C − A

2.

First Solution: Let α = A2 , β = B

2 , γ = C2 . Then 0◦ < α, β, γ < 90◦ and


α + β + γ = 90◦. By the difference-to-product formulas, we have

sin3A

2− cos

B − C

2= sin 3α − cos(β − γ )

= sin 3α − sin(α + 2γ )

= 2 cos(2α + γ ) sin(α − γ )

= −2 sin(α − β) sin(α − γ ).


sin3B

2− cos

C − A

2= −2 sin(β − α) sin(β − γ )

and

sin3C

2− cos

A − B

2= −2 sin(γ − α) sin(γ − β).

Hence it suffices to prove that

sin(α − β) sin(α − γ ) + sin(β − α) sin(β − γ ) + sin(γ − α) sin(γ − β)

≥ 0.

Note that this inequality is symmetric with respect to α, β, and γ , so wecan assume without loss of generality that 0◦ < α ≤ β ≤ γ < 90◦. Thenregrouping the terms on the left-hand side gives

sin(α − β) sin(α − γ ) + sin(γ − β)[sin(γ − α) − sin(β − α)],

which is positive because the function y = sin x is increasing for 0◦ < x <

90◦.

Note: Again the proof is similar to that of Schur’s inequality.

Second Solution: We maintain the same notation as in the first solution. Bythe addition and subtraction formulas, we have

sin 3α = sin α cos 2α + sin 2α cos α;cos(β − α) = sin(2α + γ ) = sin 2α cos γ + sin γ cos 2α;cos(β − γ ) = sin(2γ + α) = sin 2γ cos α + sin α cos 2γ ;

sin 3γ = sin γ cos 2γ + sin 2γ cos γ.


By the difference-to-product formulas, it follows that

sin 3α + sin 3γ − cos(β − α) − cos(β − γ )

= (sin α − sin γ )(cos 2α − cos 2γ )

+ (cos α − cos γ )(sin 2α − sin 2γ )

= (sin α − sin γ )(cos 2α − cos 2γ )

+ 2(cos α − cos γ ) cos(α + γ ) sin(α − γ ).

Note that sin x is increasing, and cos x and cos 2x are decreasing for 0◦ < x <

90◦. Since 0◦ < α, γ, α + γ < 90◦, each of the two products in the last sumis less than or equal to 0. Hence

sin 3α + sin 3γ − cos(β − α) − cos(β − γ ) ≤ 0.


sin 3β + sin 3α − cos(γ − β) − cos(γ − α) ≤ 0

andsin 3γ + sin 3β − cos(α − γ ) − cos(α − β) ≤ 0.

Adding the last three inequalities gives the desired result.

33. Let x1, x2, . . . , xn, n ≥ 2, be n distinct real numbers in the interval [−1, 1].Prove that

1

t1+ 1

t2+ · · · + 1

tn≥ 2n−2,

where ti =∏j �=i |xj − xi |.

Solution: Let Tn denote the nth Chebyshev polynomial. Recall that (Intro-ductory Problem 49) Tn(cos x) = cos nx and Tn is defined by the recursionTn+1(x) = 2xTn(x) − Tn−1(x), T0(x) = 1, and T1(x) = x. Therefore, theleading coefficient of Tn is 2n−1 for n ≥ 1.

Now we apply the above information to the problem at hand. We can ap-ply Lagrange’s interpolation formula to the points x1, x2, . . . , xn and thepolynomial Tn−1(x) to obtain

Tn−1(x) =n∑

k=1

Tn−1(xk)(x − x1) · · · (x − xk−1)(x − xk+1) · · · (x − xn)

(xk − x1) · · · (xk − xk−1)(xk − xk+1) · · · (xk − xn).


Equating leading coefficients, we have

2n−2 =n∑

k=1

Tn−1(xk)

(xk − x1) · · · (xk − xk−1)(xk − xk+1) · · · (xk − xn).

Set θk such that cos θk = xk . Then |Tn−1(xk)| = | cos(n−1)θk| ≤ 1. It followsthat

2n−2 ≤n∑

k=1

|Tn−1(xk)||(xk − x1) · · · (xk − xk−1)(xk − xk+1) · · · (xk − xn)|

=n∑

k=1

1

tk,

as desired.

34. [St. Petersburg 2001] Let x1, . . . , x10 be real numbers in the interval [0, π/2]such that sin2 x1 + sin2 x2 + · · · + sin2 x10 = 1. Prove that

3(sin x1 + · · · + sin x10) ≤ cos x1 + · · · + cos x10.

Solution: Because sin2 x1 + sin2 x2 + · · · + sin2 x10 = 1,

cos xi =√∑

j �=i

sin2 xj .

By the power mean inequality, for each 1 ≤ i ≤ 10,

cos xi =√∑

j �=i

sin2 xj ≥∑

j �=i sin xj

3.

Summing over all the terms cos xi gives

10∑i=1

cos xi ≥10∑i=1

∑j �=i

sin xj

3=

10∑i=1

9 · sin xi

3= 3

10∑i=1

sin xi,

as desired.


35. [IMO 2001 shortlist] Let x1, x2, . . . , xn be arbitrary real numbers. Prove theinequality

x1

1 + x21

+ x2

1 + x21 + x2

2

+ · · · + xn

1 + x21 + · · · + x2

n

<√

n.

Solution: (By Ricky Liu) We make the following substitutions: x1 = tan α1,x2 = sec α1 tan α2, and

xk = sec α1 sec α2 · · · sec αk−1 tan αk,

with −π/2 < αk < π/2, 1 ≤ k ≤ n. Note that this is always possible becausethe range of tan α is (−∞, ∞) and sec α is always nonzero. Then the kth termon the left-hand side of our inequality becomes

sec α1 · · · sec αk−1 tan αk

1 + tan2 α1 + · · · + sec2 α1 · · · sec2 αn−1 tan2 αn

= cos α1 cos α2 · cos αk sin αk.

Hence the given inequality reduces to

cos α1 sin α1 + cos α1 cos α2 sin α2 + · · · + cos α1 cos α2 · · · cos αn sin αn

<√

n;that is,

c1s1 + c1c2s2 + · · · + c1c2 · · · cnsn <√

n,

where ci = cos αi and si = sin αi for 1 ≤ i ≤ n. For 2 ≤ i ≤ n, becausec2i + s2

i = cos2 αi + sin2 αi = 1, we note that

c21c

22 · · · c2

i−1s2i + c2

1c22 · · · c2

i−1c2i = c2

1c22 · · · c2

i−1.

Therefore,

s21 + c2

1s22 + · · · + c2

1c22 · · · c2

n−2s2n−1 + c2

1c22 · · · c2

n−1 = 1. (∗)

By (∗) and Cauchy–Schwarz inequality, we obtain

c1s1 + c1c2s2 + · · · + c1c2 · · · cnsn

≤√

s21 + c2

1s22 + · · · + c2

1c22 · · · c2

n−2s2n−1 + c2

1c22 · · · c2

n−1

·√

c21 + c2

2 + · · · + c2n−1 + c2

ns2n

=√

c21 + c2

2 + · · · + c2n−1 + c2

ns2n

=√

cos2 α1 + cos2 α2 + · · · + cos2 αn−1 + cos2 αn sin2 αn

≤ √n.


Equality in the last step can hold only when

cos α1 = cos α2 = · · · = cos αn−1 = cos αn sin αn = 1,

which is impossible, because cos αn sin αn = 12 sin 2αn < 1. Therefore, we

always have strict inequality, and we are done.

36. [USAMO 1998] Let a0, a1, . . . , an be numbers in the interval(0, π

2

)such that

tan(a0 − π

4

)+ tan

(a1 − π

4

)+ · · · + tan

(an − π

4

)≥ n − 1.

Prove thattan a0 tan a1 · · · tan an ≥ nn+1.

Solution: Let bk = tan(ak − π

4

), k = 0, 1, . . . , n. It follows from the hy-

pothesis that for each k, −1 < bk < 1, and

1 + bk ≥∑

0≤��=k≤n

(1 − b�). (∗)

Applying the arithmetic–geometric means inequality to the positive realnumbers 1 − b�, � = 0, 1, . . . , k − 1, k + 1, . . . , n, we obtain

∑0≤��=k≤n

(1 − b�) ≥ n

⎛⎝ ∏0≤��=k≤n

(1 − b�)

⎞⎠1/n

. (∗∗)

From inequalities (∗) and (∗∗) it follows that

n∏k=0

(1 + bk) ≥ nn+1

(n∏

�=0

(1 − b�)n

)1/n

,

and hence thatn∏

k=0

1 + bk

1 − bk

≥ nn+1.

Because

1 + bk

1 − bk

= 1 + tan(ak − π

4

)1 − tan

(ak − π

4

) = tan[(

ak − π

4

)+ π

4

]= tan ak,

the conclusion follows.


Note: Using a similar method, one can show that

1

n − 1 + a1+ 1

n − 1 + a2+ · · · + 1

n − 1 + an

≤ 1,

where a1, a2, . . . , an are positive real numbers such that a1a2 · · · an = 1.An interesting exercise is to provide a trigonometry interpretation for the lastinequality.

37. [MOSP 2001] Find all triples of real numbers (a, b, c) such that a2 −2b2 = 1,2b2 − 3c2 = 1, and ab + bc + ca = 1.

Solution: Since a2 − 2b2 = 1, a �= 0. Since 2b2 − 3c2 = 1, b �= 0. Ifc = 0, then b = 1/

√2 and a = √

2. It is easy to check that (a, b, c) =(√2, 1/

√2, 0)

is a solution of the system. We claim that there are no other

valid triples.

We approach the problem indirectly by assuming that there exists a triple ofreal numbers (a, b, c), with abc �= 0, such that the equations hold. Withoutloss of generality, we assume that two of the numbers are positive; otherwise,we can consider the triple (−a, −b, −c). Without loss of generality, we assumethat a and b are positive. (The first two equations are independent of the signsof a, b, c, and the last equation is symmetric with respect to a, b, and c.) ByIntroductory Problem 21, we may assume that a = cot A, b = cot B, andc = cot C, with 0 < A, B < 90◦, where A, B, C are angles of a triangle. Wehave

a2 + 1 = 2(b2 + 1

)= 3(c2 + 1

).

The last equation reduces to

csc2 A = 2 csc2 B = 3 csc2 C,

or1

sin A=

√2

sin B=

√3

sin C.

By the law of sines, we conclude that the sides opposite angles A, B, C havelengths k,

√2k,

√3k, respectively, for some positive real number k. But then

triangle ABC is a right triangle with � C = 90◦, implying that c = cot C = 0,a contradiction to the assumption that c �= 0. Hence our assumption was

wrong, and (a, b, c) =(√

2, 1/√

2, 0)

is the only valid triple sought.


38. Let n be a positive integer, and let θi be angles with 0 < θi < 90◦ such that

cos2 θ1 + cos2 θ2 + · · · + cos2 θn = 1.

Prove that

tan θ1 + tan θ2 + · · · + tan θn ≥ (n − 1)(cot θ1 + cot θ2 + · · · + cot θn).

Solution: (By Tiankai Liu) By the power mean inequality, for positivenumbers x1, x2, . . . , xn, we have M−1 ≤ M1 ≤ M2; that is,

n1x1

+ 1x2

+ · · · + 1xn

≤ x1 + x2 + · · · + xn

n≤√

x21 + x2

2 + · · · + x2n

n.

For 1 ≤ i ≤ n, let cos θi = ai . Then

tan θi = sin θi

cos θi

=√

1 − cos2 θi

ai

=√

a21 + a2

2 + · · · + a2i−1 + a2

i+1 + · · · + a2n

ai

≥ a1 + a2 + · · · + ai−1 + ai+1 + · · · + an

ai

√n − 1

.

Summing the above inequalities for i from 1 to n gives

n∑i=1

tan θi ≥ 1√n − 1

n∑i=1

∑j �=i

aj

ai

= 1√n − 1

∑1≤i,j≤n

i �=j

aj

ai

, (∗)

because each ratio ai

ajappears exactly once.

On the other hand, we have

cot θi = cos θi

sin θi

= ai√1 − cos2 θi

= ai√a2

1 + a22 + · · · + a2

i−1 + a2i+1 + · · · + a2

n

≤ai

(1a1

+ 1a2

+ · · · + 1ai−1

+ 1ai+1

+ · · · + 1an

)(n − 1)

√n − 1

,


by the power mean inequality. Summing the above identities from 1 to n yields

n∑i=1

cot θi ≤ 1

(n − 1)3/2

n∑i=1

∑j �=i

ai

aj

= 1

(n − 1)3/2

∑1≤i,j≤n

i �=j

ai

aj

(∗∗)

again, because each ratio ai

ajappears once. Combining inequalities (∗) and

(∗∗) gives

√n − 1

n∑i=1

tan θi ≥∑

1≤i,j≤ni �=j

aj

ai

=∑

1≤i,j≤ni �=j

ai

aj

≥ (n − 1)3/2n∑

i=1

cot θi,


39. [Weichao Wu] One of the two inequalities

(sin x)sin x < (cos x)cos x and (sin x)sin x > (cos x)cos x

is always true for all real numbers x such that 0 < x < π4 . Identify that

inequality and prove your result.

Solution: The first inequality is true. Observe that the logarithm function isconcave down. We apply Jensen’s inequality to the points sin x < cos x <

sin x + cos x with weights λ1 = tan x and λ2 = 1 − tan x (because 0 < x <

π/4, λ1 and λ2 are positive) to obtain

log(cos x) = log[tan x sin x + (1 − tan x)(sin x + cos x)]> tan x log(sin x) + (1 − tan x) log(sin x + cos x).

Since sin x + cos x = √2 sin(x + π

4

)> 1 and tan x < 1 in the specified

interval, the second term is positive and we may drop it to obtain

log(cos x) > tan x log(sin x).

Multiplying by cos x and exponentiating gives the required inequality.

40. Let k be a positive integer. Prove that√

k + 1 − √k is not the real part of the

complex number z with zn = 1 for some positive integer n.

Note: In June 2003, this problem was first given in the training of the ChineseIMO team, and then in the MOSP. The following solution was due to AndersKaseorg, gold medalist in the 44th IMO in July 2003 in Tokyo, Japan.


Solution: Assume to the contrary that α = √k + 1 − √

k is the real part ofsome complex number z with zn = 1 for some positive integer n. Because z

is an nth root of unity, it can be written as cos 2πjn

+ i sin 2πjn

for some integer

j with 0 ≤ j ≤ n − 1. Thus, α = cos 2πjn

.

Let Tn(x) be the nth Chebyshev polynomial; that is, T0(x) = 1, T1(x) = x,and Ti+1 = 2xTi(x)−Ti−1(x) for i ≥ 1. Then Tn(cos θ) = cos(nθ), implyingthat Tn(α) = cos(2πj) = 1.

Let β = √k + 1 + √

k. Note that αβ = 1 and α + β = 2√

k + 1, and soα2 + β2 = (α + β)2 − 2αβ = 4k + 2. Thus ±α and ±β are the roots of thepolynomial

P(x) = (x − α)(x + α)(x − β)(x + β) =(x2 − α2

) (x − β2

)= x4 − (4k + 2)x + 1.

Let Q(x) be the minimal polynomial for α. If neither β nor −β is a root ofQ(x), then Q(x) must divide

(x − α)(x + α) = x2 − [2k + 1 − 2√

k(k + 1)],and so k(k + 1) must be a perfect square, which is impossible because k2 <

k(k + 1) < (k + 1)2. Therefore, either Q(β) or Q(−β) = 0 or both. We saythat Q(β ′) = 0, with either β ′ = β or β ′ = −β.

Because α is a root of Tn(x) − 1, Q(x) divides Tn(x) − 1 and β ′ is a root ofTn(x)−1. However, by Introductory Problem 49(f), the roots ofTn(x)−1 are allin the interval [−1, 1] and |β ′| = √

k + 1+√k > 1, which is a contradiction.

Therefore, our original assumption was wrong, and√

k + 1 − √k is not the

real part of any nth root of unity.

41. Let A1A2A3 be an acute-angled triangle. Points B1, B2, B3 are on sides A2A3,

A3A1, A1A2, respectively. Prove that

2(b1 cos A1 + b2 cos A2 + b3 cos A3) ≥ a1 cos A1 + a2 cos A2 + a3 cos A3,

where ai = |Ai+1Ai+2| and bi = |Bi+1Bi+2| for i = 1, 2, 3 (with indicestaken modulo 3; that is, xi+3 = xi).

Solution: As shown in Figure 5.10, let |BiAi+1| = si and |BiAi+2| = ti ,i = 1, 2, 3. Then ai = si + ti . Our approach here is similar to that of AdvancedProblem 26. Let A1 = � A1, A2 = � A2, and A3 = � A3.


A1

B1A2 A3

B2

B3

E F

s1

s2

s3

t1

t2

t3

Figure 5.10.

Note that segment EF , the projection of segment B2B3 onto the line A2A3,has length a1 − t3 cos A2 − s2 cos A3, and so

b1 ≥ a1 − t3 cos A2 − s2 cos A3.

Because 0 < A1 < 90◦, we know that

b1 cos A1 ≥ a1 cos A1 − t3 cos A2 cos A1 − s2 cos A3 cos A1.

Likewise, we find that

b2 cos A2 ≥ a2 cos A2 − t1 cos A3 cos A2 − s3 cos A1 cos A2

and

b3 cos A3 ≥ a3 cos A3 − t2 cos A1 cos A3 − s1 cos A2 cos A3.

Adding the last three inequalities, we observe that

3∑i=1

bi cos Ai ≥3∑

i=1

ai(cos Ai − cos Ai+1Ai+2).


23∑

i=1

ai(cos Ai − cos Ai+1Ai+2) ≥3∑

i=1

ai cos Ai,

or3∑

i=1

ai(cos Ai − 2 cos Ai+1Ai+2) ≥ 0.


Applying the law of sines to triangle A1A2A3, the last inequality reduces to

3∑i=1

sin Ai(cos Ai − 2 cos Ai+1Ai+2) ≥ 0,

which follows directly from the next Lemma.

Lemma Let ABC be a triangle. Then the Cyclic Sum∑cyc

sin A(cos A − 2 cos B cos C) = 0.

Proof: By the double-angle formulas, it suffices to show that∑cyc

sin 2A = 2∑cyc

sin A cos A = 4∑cyc

sin A cos B cos C.

Applying the addition and subtraction formulas gives

sin A cos B cos C + sin B cos C cos A

= cos C(sin A cos B + sin B cos A)

= cos C sin(A + B) = cos C sin C.

Hence

4∑cyc

sin A cos B cos C

= 2∑cyc

(sin A cos B cos C + sin B cos C cos A)

= 2∑cyc

cos C sin C =∑cyc

sin 2C,

as desired.

42. Let ABC be a triangle. Let x, y, and z be real numbers, and let n be a positiveinteger. Prove the following four inequalities.

(a) [D. Barrow] x2 + y2 + z2 ≥ 2yz cos A + 2zx cos B + 2xy cos C.

(b) [J. Wolstenholme]

x2 + y2 + z2 ≥ 2(−1)n+1(yz cos nA + zx cos nB + xy cos nC).


(c) [O. Bottema] yza2 + zxb2 + xyc2 ≤ R2(x + y + z)2.

(d) [A. Oppenheim] xa2 + yb2 + zc2 ≥ 4[ABC]√xy + yz + zx.

Note: These are very powerful inequalities, because x, y, z can be arbitraryreal numbers. By the same token, however, they are not easy to apply.

Solution: It is clear that part (a) is a special case of part (b) by setting n = 1.Parts (c) and (d) are two applications of part (b). Hence we prove only parts(b), (c), and (d).

(b) Rewrite the desired inequality as

x2 + 2x(−1)n(z cos nB +y cos nC)+y2 + z2 + 2(−1)nyz cos nA ≥ 0.

Completing the square for x2 + 2x(−1)n(z cos nB + y cos C) gives[x2 + (−1)n(z cos nB + y cos nC)

]2 + y2 + z2 + 2(−1)nyz cos nA

≥ (z cos nB + y cos nC)2

= z2 cos2 nB + y2 cos2 nC + 2yz cos nB cos nC.


y2 + z2 + 2(−1)nyz cos nA ≥ z2 cos2 nB + y2 cos2 nC

+ 2yz cos nB cos nC,

or

y2 sin2 nC + z2 sin2 nB + 2yz[(−1)n cos nA − cos nB cos nC

] ≥ 0.

(∗)

If n = 2k is even, then nA + nB + nC = 2kπ , and so cos nA =cos(nB +nC) = cos nB cos nC − sin nB sin nC. The desired inequality(∗) reduces to

y2 sin2 nC + z2 sin2 nB − 2yz sin nB sin nC

= (y sin nC − z sin nB)2 ≥ 0,

which is evident.

If n = 2k + 1 is even, then nA + nB + nC = (2k + 1)π , and socos nA = − cos(nB + nC) = − cos nB cos nC + sin nB sin nC. Thedesired inequality (∗) reduces to


y2 sin2 nC + z2 sin2 nB − 2yz sin nB sin nC

= (y sin nC − z sin nB)2 ≥ 0,

which is evident.From the above proof, we note that the equality case of the desiredinequality holds only if (y sin nC −z sin nB)2 ≥ 0, that is, if y sin nC =z sin nB, or y

sin nB= z

sin nC. By symmetry, the equality case holds only

ifx

sin nA= y

sin nB= z

sin nC.

It is also easy to check that the above condition is sufficient for theequality case to hold.

(c) The extended law of sines gives that aR

= 2 sin A and its analogousforms for b

Rand c

R. Dividing both sides of the desired inequalities by R2

and expanding the resulting right-hand side yields

4(yz sin2 A + zx sin2 B + xy sin2 C)

≤ x2 + y2 + z2 + 2(xy + yz + zx),

or, by the double-angle formulas,

x2 + y2 + z2

≥ 2[yz(2 sin2 A − 1) + zx(2 sin2 B − 1) + xy(2 sin2 C − 1

]= −2(yz cos 2A + zx cos 2B + xy cos 2C),

which is part (b) by setting n = 2.By the argument at the end of the proof of (b), we conclude that theequality case of the desired inequality holds if and only if

x

sin 2A= y

sin 2B= z

sin 2C.

(d) Setting x = xa2, y = yb2, and z = zc2 in part (c) gives

a2b2c2(xy + yz + zx) ≤ R2(xa2 + yb2 + zc2

)2,

or

16R2[ABC](xy + yz + zx) ≤ R2(xa2 + yb2 + zc2

)2,

by Introductory Problem 25(a). Dividing both sides of the last inequalityby R2 and taking square roots yields the desired result.


By the argument at the end of the proof of (b), we conclude that theequality case of the desired inequality holds if and only if

xa2

sin 2A= yb2

sin 2B= zc2

sin 2C,

orxa

cos A= yb

cos B= zc

sin C,

by the double-angle formulas and the law of sines. By the law of cosines,we have

a

cos A= 2abc

b2 + c2 − a2

and its analogous forms for bcos B

and ccos C

. Therefore, the equality caseholds if and only if

x

b2 + c2 − a2 = y

c2 + a2 − b2 = z

a2 + b2 − c2 .

Note: The approach of completing the squares, shown in the proof of (b),is rather tricky. We can see this approach from another angle. Consider thequadratic function

f (x) = x2 − 2x(z cos B + y cos C) + y2 + z2 − 2yz cos A.

Its discriminant is

= 4(z cos B + y cos C)2 − 4(y2 + z2 − 2yz cos A)

= 4[z2 cos2 B − z2 + 2yz(cos A + cos B cos C) + y2 cos2 C − y2

]= 4[−z2 sin2 B + 2yz(− cos(B + C) + cos B cos C) − y2 sin2 C

]= 4[−z2 sin2 B + 2yz sin B sin C − y2 sin2 C

]= −4(z sin B − y sin C)2 ≤ 0.

For fixed y and z and large x, f (x) is certainly positive. Hence f (x) ≥ 0 forall x, establishing (a). This method can be easily generalized to prove (b). Weleave the generalization to the reader as an exercise.

43. [USAMO 2004] A circle ω is inscribed in a quadrilateral ABCD. Let I be thecenter of ω. Suppose that


(|AI | + |DI |)2 + (|BI | + |CI |)2 = (|AB| + |CD|)2.

Prove that ABCD is an isosceles trapezoid.

Note: We introduce two trigonometric solutions and a synthetic-geometricsolution. The first solution, by Oleg Golberg, is very technical. The second so-lution, by Tiankai Liu and Tony Zhang, reveals more geometrical backgroundin their computations. This is by far the most challenging problem in the US-AMO 2004. There were only four complete solutions. The fourth student isJacob Tsimerman, from Canada. Evidently, these four students placed top fourin the contest. There were nine IMO gold medals won by these four students,with each of Oleg and Tiankai winning three, Jacob two, and Tony one. Olegwon his first two representing Russia, and the third representing the UnitedStates. Jacob is one of only four students who achieved a perfect score at theIMO 2004 in Athens, Greece.

The key is to recognize that the given identity is a combination of equalitycases of certain inequalities. By equal tangents, we have |AB| + |CD| =|AD| + |BC| if and only if ABCD has an incenter. We will prove that for aconvex quadrilateral ABCD with incenter I , then

(|AI |+|DI |)2 +(|BI |+|CI |)2 ≤ (|AB|+|CD|)2 = (|AD|+|BC|)2. (∗)

Equality holds if and only if AD ‖ BC and |AB| = |CD|. Without loss ofgenerality, we may assume that the inradius of ABCD is 1.

A

B C

D

I

A1

B1

C1

x

y z

w

D1

Figure 5.11.

First Solution: As shown in Figure 5.11, let A1, B1, C1, and D1 be the pointsof tangency. Because circle ω is inscribed in ABCD, we can set � D1IA =� AIA1 = x, � A1IB = � BIB1 = y, � B1IC = � CIC1 = z, � C1ID =� DID1 = w, and x + y + z + w = 180◦, or x + w = 180◦ − (y + z),


with 0◦ < x, y, z, w < 90◦. Then |AI | = sec x, |BI | = sec y, |CI | = sec z,|DI | = sec w, |AD| = |AD1| + |D1D| = tan x + tan w, and |BC| =|BB1| + |B1C| = tan y + tan z. Inequality (∗) becomes

(sec x + sec w)2 + (sec y + sec z)2 ≤ (tan x + tan y + tan z + tan w)2.

Expanding both sides of the above inequality and applying the identity sec2 x =1 + tan2 x gives

4 + 2(sec x sec w + sec y sec z)

≤ 2 tan x tan y + 2 tan x tan z + 2 tan x tan w

+ 2 tan y tan z + 2 tan y tan w + 2 tan z tan w,

or

2 + sec x sec w + sec y sec z

≤ tan x tan w + tan y tan z + (tan x + tan w)(tan y + tan z).

Note that by the addition and subtraction formulas,

1 − tan x tan w = cos x cos w − sin x sin w

cos x cos w= cos(x + w)

cos x cos w.

Hence,

1 − tan x tan w + sec x sec w = 1 + cos(x + w)

cos x cos w.

Similarly,

1 − tan y tan z + sec y sec z = 1 + cos(y + z)

cos y cos z.

Adding the last two equations gives

2 + sec x sec w + sec y sec z − tan x tan w − tan y tan z

= 1 + cos(x + w)

cos x cos w+ 1 + cos(y + z)

cos y cos z.


1 + cos(x + w)

cos x cos w+ 1 + cos(y + z)

cos y cos z≤ (tan x + tan w)(tan y + tan z),

ors + t ≤ (tan x + tan w)(tan y + tan z),

after setting s = 1+cos(x+w)cos x cos w

and t = 1+cos(y+z)cos x cos w

. By the addition and subtrac-tion formulas, we have

tan x + tan w = sin x cos w + cos x sin w

cos x cos w= sin(x + w)

cos x cos w.


Similarly,

tan y + tan z = sin(y + z)

cos y cos z= sin(x + w)

cos y cos z,

because x + w = 180◦ − (y + z). It follows that

(tan x + tan w)(tan y + tan z)

= sin2(x + w)

cos x cos y cos z cos w= 1 − cos2(x + w)

cos x cos y cos z cos w

= [1 − cos(x + w)][1 + cos(x + w)]cos x cos y cos z cos w

= [1 + cos(y + z)][1 + cos(x + w)]cos x cos y cos z cos w

= st.

The desired inequality becomes s+t ≤ st , or (1−s)(1−t) = 1−s−t+st ≥ 1.It suffices to show that 1 − s ≥ 1 and 1 − t ≥ 1. By symmetry, we have onlyto show that 1 − s ≥ 1; that is,

1 + cos(x + w)

cos x cos w≥ 2.

Multiplying both sides of the inequality by cos x cos w and applying the addi-tion and subtraction formulas gives

1 + cos x cos w − sin x sin w ≥ 2 cos x cos w,

or 1 ≥ cos x cos w + sin x sin w = cos(x − w), which is evident. Equalityholds if and only if x = w. Therefore, inequality (∗) is true, with equality ifand only if x = w and y = z, which happens precisely when AD ‖ BC and|AB| = |CD|, as was to be shown.

Second Solution: We maintain the same notation as in the first solution.Applying the law of cosines to triangles ADI and BCI gives

|AI |2 + |DI |2 = 2 cos(x + w)|AI | · |DI | + |AD|2

and|BI |2 + |CI |2 = 2 cos(y + z)|BI | · |CI | + |BC|2.

Adding the last two equations and completing squares gives

(|AI | + |DI |)2 + (|BI | + |CI |)2 + 2|AD| · |BC|= 2 cos(x + w)|AI | · |DI | + 2 cos(y + z)|BI | · |CI |

+ 2|AI | · |DI | + 2|BI | · |CI | + (|AD| + |BC|)2.


Hence, establishing inequality (∗) is equivalent to establishing the inequality

[1 + cos(x + w)]|AI | · |DI | + [1 + cos(y + z)]|BI | · |CI | ≤ |AD| · |BC|.Since 2[ADI ] = |AD|·|ID1| = |AI |·|DI | sin(x+w), |AI |·|DI | = |AD|

sin(x+w).

Similarly, |BI | · |CI | = |BC|sin(y+z)

. Because x + w = 180◦ − (y + z), we havesin(x + w) = sin(y + z) and cos(x + w) = − cos(x + w). Plugging all theabove information back into the last inequality yields

1 + cos(x + w)

sin(x + w)· |AD| + 1 − cos(x + w)

sin(x + w)· |BC| ≤ |AD| · |BC|,

or1 + cos(x + w)

|BC| + 1 − cos(x + w)

|AD| ≤ sin(x + w). (∗∗)

Note that by the addition and subtraction, the product-to-sum, and the double-angle formulas, we have

|AD| = |AD1| + |D1D| = tan x + tan w = sin x

cos x+ sin w

cos w

= sin x cos w + cos x sin w

cos x cos w= sin(x + w)

cos x cos w= 2 sin(x + w)

2 cos x cos w

= 4 sin x+w2 cos x+w

2

cos(x + w) + cos(x − w)≥ 4 sin x+w

2 cos x+w2

cos(x + w) + 1

= 4 sin x+w2 cos x+w

2

2 cos2 x+w2

= 2 tanx + w

2.

Equality holds if and only if cos(x − w) = 1, that is, if x = w. (This step canbe done easily by applying Jensen’s inequality, using the fact y = tan x isconvex for 0◦ < x < 90◦.) Consequently, by the double-angle formulas,

1 − cos(x + w)

|AD| ≤ 2 sin2 x+w2

2 tan x+w2

= sinx + w

2cos

x + w

2

= sin(x + w)

2.


1 + cos(x + w)

|BC| = 1 − sin(y + z)

|BC| ≤ sin(y + z)

2= sin(x + w)

2.

Adding the last two inequalities gives the desired inequality (∗∗). Equalityholds if and only if x = w and y = z, which happens precisely when AD ‖ BC

and |AB| = |CD|, as was to be shown.


Third Solution: Because the circle ω is inscribed in ABCD, as shown inFigure 5.12, we can set � DAI = � IAB = a, � ABI = � IBC = b, � BCI =� ICD = c, � CDI = � IDA = d , and a + b + c + d = 180◦. Our proof isbased on the following key Lemma.

Lemma: If a circle ω, centered at I , is inscribed in a quadrilateral ABCD,then

|BI |2 + |AI ||DI | · |BI | · |CI | = |AB| · |BC|. (†)

A

B C

D

I

a aa

b

b

b c

cc

dd

d

Figure 5.12.

Proof: Construct a point P outside of the quadrilateral such that triangleABP is similar to triangle DCI . We obtain

� PAI + � PBI = � PAB + � BAI + � PBA + � ABI

= � IDC + a + � ICD + b

= a + b + c + d = 180◦,

implying that the quadrilateral PAIB is cyclic. By Ptolemy’s theorem, wehave |AI | · |BP | + |BI | · |AP | = |AB| · |IP |, or

|BP | · |AI ||IP | + |BI | · |AP |

|IP | = |AB|. (††)

Because PAIB is cyclic, it is not difficult to see that, as indicated in thefigure, � IPB = � IAB = a, � API = � ABI = b, � AIP = � ABP = c,and � PIB = � PAB = d. Note that triangles AIP and ICB are similar,implying that

|AI ||IP | = |IC|

|CB| and|AP ||IP | = |IB|

|CB| .Substituting the above equalities into the identity (††), we arrive at

|BP | · |CI ||BC| + |BI |2

|BC| = |AB|,


or|BP | · |CI | + |BI |2 = |AB| · |BC|. (†††)

Note also that triangleBIP and triangle IDA are similar, implying that |BP ||BI | =

|IA||ID| , or

|BP | = |AI ||ID| · |IB|.

Substituting the above identity back into (†††) gives the desired relation (†),establishing the Lemma.

Now we prove our main result. By the Lemma and symmetry, we have

|CI |2 + |DI ||AI | · |BI | · |CI | = |CD| · |BC|. (‡)

Adding the two identities (†) and (‡) gives

|BI |2 + |CI |2 +( |AI |

|DI | + |DI ||AI |)

|BI | · |CI | = |BC|(|AB| + |CD|).

By the arithmetic–geometric means inequality, we have |AI ||DI | + |DI |

|AI | ≥ 2.Thus,

|BC|(|AB| + |CD|) ≥ |IB|2 + |IC|2 + 2|IB| · |IC| = (|BI | + |CI |)2,

where equality holds if and only if |AI | = |DI |. Likewise, we have

|AD|(|AB| + |CD|) ≥ (|AI | + |DI |)2,

where equality holds if and only if |BI | = |CI |. Adding the last two identitiesgives the desired inequality (∗).

By the given condition in the problem, all the equalities in the above discussionmust hold; that is, |AI | = |DI | and |BI | = |CI |. Consequently, we havea = d, b = c, and so � DAB + � ABC = 2a + 2b = 180◦, implyingthat AD ‖ BC. It is not difficult to see that triangle AIB and triangle DIC

are congruent, implying that |AB| = |CD|. Thus, ABCD is an isoscelestrapezoid.

44. [USAMO 2001] Let a, b, and c be nonnegative real numbers such that

a2 + b2 + c2 + abc = 4.

Prove that0 ≤ ab + bc + ca − abc ≤ 2.


The proof of the lower bound is rather simple. From the given condition, atleast one of a, b, and c does not exceed 1, say a ≤ 1. Then

ab + bc + ca − abc = a(b + c) + bc(1 − a) ≥ 0.

To obtain equality, we have a(b+c) = bc(1−a) = 0. If a = 1, then b+c = 0or b = c = 0, which contradicts the fact that a2 + b2 + c2 + abc = 4. Hence1 − a �= 0, and only one of b and c is 0. Without loss of generality, say b = 0.Therefore b + c > 0 and a = 0. Plugging a = b = 0 back into the givencondition, we get c = 2. By permutation, the lower bound holds if and only if(a, b, c) is one of the triples (2, 0, 0), (0, 2, 0), and (0, 0, 2). We next presentthree proofs of the upper bound.

First Solution: Based on Introductory Problem 22, we set a = 2 sin A2 ,

b = 2 sin B2 , and c = 2 sin C

2 , where ABC is a triangle. We have

ab = 4 sinA

2sin

B

2= 2

√sin A tan

A

2sin B tan

B

2

= 2

√sin A tan

B

2sin B tan

A

2.

By the arithmetic–geometric means inequality, this is at most

sin A tanB

2+ sin B tan

A

2

= sin A cotA + C

2+ sin B cot

B + C

2.

Likewise,

bc ≤ sin B cotB + A

2+ sin C cot

C + A

2,

ca ≤ sin C cotC + B

2+ sin A cot

A + B

2.

Therefore, by the sum-to-product, product-to-sum, and the double-angleformulas, we have


ab + bc + ca

≤ (sin A + sin B) cotA + B

2+ (sin B + sin C) cot

B + C

2

+ (sin C + sin A) cotC + A

2

= 2 cosA − B

2cos

A + B

2+ 2 cos

B − C

2cos

B + C

2

+ 2 cosC − A

2cos

C + A

2= 2(cos A + cos B + cos C)

= 6 − 4

(sin2 A

2+ sin2 B

2+ sin2 C

2

)= 6 −

(a2 + b2 + c2

).

Using the given equality, this last quantity equals 2 + abc. It follows that

ab + bc + ca ≤ 2 + abc,

as desired.

Second Solution: Clearly, 0 ≤ a, b, c ≤ 2. In the light of IntroductoryProblem 24(d), we can set a = 2 cos A, b = 2 cos B, and c = 2 cos C, whereABC is an acute triangle. Either two of A, B, and C are at least 60◦, or twoof A, B, and C are at most 60◦. Without loss of generality, assume that A andB have this property.

With these trigonometric substitutions, we find that the desired inequality isequivalent to

2(cos A cos B + cos B cos C + cos C cos A) ≤ 1 + 4 cos A cos B cos C,

or,

2(cos A cos B + cos B cos C + cos C cos A)

≤ 3 − 2(cos2 A + cos2 B + cos2 C).

Hence, by the double-angle formulas, it suffices to prove that

cos 2A + cos 2B + cos 2C

+ 2(cos A cos B + 2 cos B cos C + cos C cos A) ≤ 0.


By the sum-to-product and double-angle formulas, the sum of the first threeterms in this inequality is

cos 2A + cos 2B + cos 2C

= 2 cos (A + B) cos (A − B) + 2 cos2 (A + B) − 1

= 2 cos (A + B)[cos (A − B) + cos (A + B)] − 1

= 4 cos (A + B) cos A cos B − 1,

while the remaining terms equal

2 cos A cos B + 2 cos C(cos A + cos B)

= cos (A + B) + cos (A − B) − 2 cos (A + B)(cos A + cos B),

by the product-to-sum formulas. Hence, it suffices to prove that

cos (A + B)[4 cos A cos B + 1 − 2 cos A − 2 cos B] + cos (A − B) ≤ 1,

or− cos C(1 − 2 cos A)(1 − 2 cos B) + cos (A − B) ≤ 1. (∗)

We consider the following cases:

(i) At least one angle is 60◦. If A or B equals 60◦, then we may assume,without loss of generality, that A = 60◦. If C = 60◦, then because eitherA, B ≥ 60◦ or A, B ≤ 60◦, we must actually have A = B = 60◦, inwhich case equality holds. In either case, we may assume A = 60◦. Then(∗) becomes cos (A − B) ≤ 1, which is always true, and equality holdsif and only if A = B = C = 60◦, that is, if and only if a = b = c = 1.

(ii) No angle equals 60◦. Because either A, B ≥ 60◦ or A, B ≤ 60◦, we have(1 − 2 cos A)(1 − 2 cos B) > 0. Since cos C ≥ 0 and cos (A − B) ≤ 1,(∗) is always true. Equality holds when cos C = 0 and cos (A − B) = 1.

This holds exactly when A = B = 45◦ and C = 90◦; that is, whena = b = √

2 and c = 0.

Third Solution: The problem also admits the following clever purely alge-braic method, which is due to Oaz Nir and Richard Stong, independently.

Either two of a, b, c are less than or equal to 1, or two are greater than or equalto 1. Assume that b and c have this property. Then

b + c − bc = 1 − (1 − b)(1 − c) ≤ 1. (†)


Viewing the given equality as a quadratic equation in a and solving for a yields

a =−bc ±

√b2c2 − 4

(b2 + c2

)+ 16

2.

Note that

b2c2 − 4(b2 + c2) + 16 ≤ b2c2 − 8bc + 16 = (4 − bc)2.

For the given equality to hold, we must have b, c ≤ 2, so that 4 − bc ≥ 0.

Hence,

a ≤ −bc + |4 − bc|2

= −bc + 4 − bc

2= 2 − bc,

or2 − bc ≥ a. (‡)

Combining the inequalities (†) and (‡) gives

2 − bc = (2 − bc) · 1 ≥ a(b + c − bc) = ab + ac − abc,

or ab + ac + bc − abc ≤ 2, as desired.

45. [Gabriel Dospinescu and Dung Tran Nam] Let s, t, u, v be numbers in theinterval

(0, π

2

)with s + t + u + v = π . Prove that

√2 sin s − 1

cos s+

√2 sin t − 1

cos t+

√2 sin u − 1

cos u+

√2 sin v − 1

cos v≥ 0.

Solution: Set a = tan s, b = tan t , c = tan u, and d = tan v. Then a, b, c, d

are positive real numbers. Because s + t + u + v = π , it follows that tan(s +t) + tan(u + v) = 0; that is,

a + b

1 − ab+ c + d

1 − cd= 0,

by the addition and subtraction formulas. Multiplying both sides of the lastequation by (1 − ab)(1 − cd) yields

(a + b)(1 − cd) + (c + d)(1 − ab) = 0,

ora + b + c + d = abc + bcd + cda + dab.


Consequently, we obtain

(a + b)(a + c)(a + d) = a2(a + b + c + d) + abc + bcd + cda + dab

= (a2 + 1)(a + b + c + d),

ora2 + 1

a + b= (a + c)(a + d)

a + b + c + d

and its analogous forms. Hence

a2 + 1

a + b+ b2 + 1

b + c+ c2 + 1

c + d+ d2 + 1

d + a

= (a + c)(a + d) + (b + d)(b + a) + (c + a)(c + b) + (d + b)(d + c)

a + b + c + d

= a2 + b2 + c2 + d2 + 2(ab + ac + ad + bc + bd + cd)

a + b + c + d

= a + b + c + d.

By Cauchy–Schwarz inequality, we have

2(a + b + c + d)2

= 2(a + b + c + d)

(a2 + 1

a + b+ b2 + 1

b + c+ c2 + 1

c + d+ d2 + 1

d + a

)= [(a + b) + (b + c) + (c + d) + (d + a)]

×(

a2 + 1

a + b+ b2 + 1

b + c+ c2 + 1

c + d+ d2 + 1

d + a

)≥(√

a2 + 1 +√

b2 + 1 +√

c2 + 1 +√

d2 + 1)2

,

or √a2 + 1 +

√b2 + 1 +

√c2 + 1 +

√d2 + 1 ≤ √

2(a + b + c + d).

The least inequality is equivalent to

1

cos s+ 1

cos t+ 1

cos u+ 1

cos v≤ √

2

(sin s

cos s+ sin t

cos t+ sin u

cos u+ sin v

cos v

),

from which the desired inequality follows.


46. [USAMO 1995] Suppose a calculator is broken and the only keys that still workare the sin, cos, tan, sin−1, cos−1, and tan−1 buttons. The display initiallyshows 0. Given any positive rational number q, show that we can get q toappear on the display panel of the calculator by pressing some finite sequenceof buttons. Assume that the calculator does real-number calculations withinfinite precision, and that all functions are in terms of radians.

Solution: Because cos−1 sin θ = π2 − θ and tan

(π2 − θ

) = 1tan θ

for 0 < θ <π2 , we have for any x > 0,

tan cos−1 sin tan−1 x = tan(π

2− tan−1 x

)= 1

x. (∗)

Also, for x ≥ 0,

cos tan−1 √x = 1√

x + 1,

so by (∗),tan cos−1 sin tan−1 cos tan−1 √

x = √x + 1. (∗∗)

By induction on the denominator of r , we now prove that√

r , for every non-negative rational number r , can be obtained by using the operations

√x �→ √

x + 1 and x �→ 1

x.

If the denominator is 1, we can obtain√

r , for every nonnegative integer r , byrepeated application of

√x �→ √

x + 1. Now assume that we can get√

r forall rational numbers r with denominator up to n. In particular, we can get anyof √

n + 1

1,

√n + 1

2, . . . ,

√n + 1

n,

so we can also get √1

n + 1,

√2

n + 1, . . . ,

√n

n + 1,

and√

r , for any positive r of exact denominator n + 1, can be obtained byrepeatedly applying

√x �→ √

x + 1.

Thus for any positive rational number r, we can obtain√

r . In particular, wecan obtain

√q2 = q.


47. [China 2003, byYumin Huang] Let n be a fixed positive integer. Determine thesmallest positive real number λ such that for any θ1, θ2, . . . , θn in the interval(0, π

2

), if

tan θ1 tan θ2 · · · tan θn = 2n/2,

thencos θ1 + cos θ2 + · · · + cos θn ≤ λ.

Solution: The answer is

λ =

⎧⎪⎨⎪⎩√

33 , n = 1;

2√

33 , n = 2;

n − 1, n ≥ 3.

The case n = 1 is trivial. If n = 2, we claim that

cos θ1 + cos θ2 ≤ 2√

3

3,

with equality if and only if θ1 = θ2 = tan−1√

2. It suffices to show that

cos2 θ1 + cos2 θ2 + 2 cos θ1 cos θ2 ≤ 4

3,

or

1

1 + tan2 θ1+ 1

1 + tan2 θ1+ 2

√1(

1 + tan2 θ1) (

1 + tan2 θ2) ≤ 4

3.

Because tan θ1 tan θ2 = 2,(1 + tan2 θ1

) (1 + tan2 θ2

)= 5 + tan2 θ1 + tan2 θ2.

By setting tan2 θ1 + tan θ22 = x, the last inequality becomes

2 + x

5 + x+ 2

√1

5 + x≤ 4

3,

or

2

√1

5 + x≤ 14 + x

3(5 + x).

Squaring both sides and clearing denominators, we get 36(5 + x) ≤ 196 +28x + x2, that is, 0 ≤ x2 − 8x + 16 = (x − 4)2. This establishes our claim.


Now assume that n ≥ 3. We claim that λ = n − 1. Note that λ ≥ n − 1; bysetting θ2 = θ3 = · · · = θn = θ and letting θ → 0, we find that θ1 → π

2 , andso the left-hand side of the desired inequality approaches n − 1. It suffices toshow that

cos θ1 + cos θ2 + · · · + cos θn ≤ n − 1.

Without loss of generality, assume that θ1 ≥ θ2 ≥ · · · ≥ θn. Then

tan θ1 tan θ2 tan θ3 ≥ 2√

2.


cos θ1 + cos θ2 + cos θ3 < 2. (∗)

Because√

1 − x2 ≤ 1 − 12x2, cos θi =

√1 − sin2 θi < 1 − 1

2 sin2 θi . Conse-quently, by the arithmetic–geometric means inequality,

cos θ2 + cos θ3 < 2 − 1

2

(sin2 θ2 + sin2 θ3

)≤ 2 − sin θ2 sin θ3.

Because

tan2 θ1 ≥ 8

tan2 θ2 tan2 θ3,

we have

sec2 θ1 ≥ 8 + tan2 θ2 tan2 θ3

tan2 θ2 tan2 θ3,

or

cos θ1 ≤ tan θ2 tan θ3√8 + tan2 θ2 tan2 θ3

= sin θ2 sin θ3√8 cos2 θ2 cos2 θ3 + sin2 θ2 sin2 θ3

.

It follows that

cos θ1 + cos θ2 + cos θ3

< 2 − sin θ2 sin θ3

[1 − 1

8 cos2 θ2 cos2 θ3 + sin2 θ2 sin2 θ3

].

It is clear that the equality

8 cos2 θ2 cos2 θ3 + sin2 θ2 sin2 θ3 ≥ 1, (∗∗)

establishes the desired inequality (∗). Inequality (∗∗) is equivalent to

8 + tan2 θ2 tan2 θ3 ≥ (1 + tan2 θ2)(1 + tan2 θ3),


or7 ≥ tan2 θ2 + tan2 θ3.

Thus if tan2 θ2 + tan2 θ3 ≤ 7, then inequality (∗) holds, and we are done.

Assume that tan2 θ2 + tan2 θ3 > 7. Then tan2 θ1 ≥ tan2 θ2 ≥ 72 . Then

cos θ1 ≤ cos θ2 = 1√1 + tan2 θ2

≤√

2

3,

implying that

cos θ1 + cos θ2 + cos θ3 ≤ 2√

2

3+ 1 < 2,

establishing (∗) again.

Therefore, inequality (∗) is true, as desired.

48. Let ABC be an acute triangle. Prove that

(sin 2B + sin 2C)2 sin A + (sin 2C + sin 2A)2 sin B

+ (sin 2A + sin 2B)2 sin C ≤ 12 sin A sin B sin C.

First Solution: Applying the addition and subtraction formulas gives

(sin 2B + sin 2C)2 sin A = 4 sin2(B + C) cos2(B − C) sin A

= 4 sin3 A cos2(B − C),

because A + B + C = 180◦. Hence it suffices to show that the cyclic sum∑cyc

sin3 A cos2(B − C)

is less than or equal to 3 sin A sin B sin C, which follows from∑cyc

4 sin3 A cos(B − C) = 12 sin A sin B sin C.

Indeed, we have

4 sin3 A cos(B − C)

= 4 sin2 A sin(B + C) cos(B − C)

= 2 sin2 A(sin 2B + sin 2C)

= (1 − cos 2A)(sin 2B + sin 2C)

= (sin 2B + sin 2C) − sin 2B cos 2A − sin 2C cos 2A.


It follows that∑cyc

4 sin3 A cos(B − C)

=∑cyc

(sin 2B + sin 2C) −∑cyc

sin 2B cos 2A −∑cyc

sin 2C cos 2A

= 2∑cyc

sin 2A −∑cyc

sin 2B cos 2A −∑cyc

sin 2A cos 2B

= 2∑cyc

sin 2A −∑cyc

(sin 2B cos 2A + sin 2A cos 2B)

= 2∑cyc

sin 2A −∑cyc

sin(2B + 2A)

= 2∑cyc

sin 2A +∑cyc

sin 2C

= 3(sin 2A + sin 2B + sin 2C)

= 12 sin A sin B sin C,

by Introductory Problem 24(a). Equality holds if and only if cos(A − B) =cos(B − C) = cos(C − A) = 1, that is, if and only if triangle ABC isequilateral.

Note: Enlarging sin3 A cos2(B −C) to sin3 A cos(B −C) is a very clever butsomewhat tricky idea. The following more geometric approach reveals moreof the motivation behind the problem. Please note the last step in the proof ofthe Lemma below.

Second Solution: We can rewrite the desired inequality as∑cyc

(sin 2B + sin 2C)2 sin A ≤ 12 sin A sin B sin C.

By the extended law of sines, we have c = 2R sin C, a = 2R sin A, andb = 2R sin B. Hence

12R2 sin A sin B sin C = 3ab sin C = 6[ABC].It suffices to show that

R2∑cyc

(sin 2B + sin 2C)2 sin A ≤ 6[ABC]. (∗)


We establish the following Lemma.

Lemma Let AD, BE, CF be the altitudes of acute triangle ABC, withD, E, F on sides BC, CA, AB, respectively. Then

|DE| + |DF | ≤ |BC|.Equality holds if and only if |AB| = |AC|.Proof: We consider Figure 5.13. Because � CFA = � CDA = 90◦, quadri-lateral AFDC is cyclic, and so � FDB = � BAC = � CAB and � BFD =� BCA = � BCA. Hence triangles BDF and BAC are similar, so

|DF ||AC| = |BF |

|BC| = cos B,

or (by the double-angle formula)

|DF | = b cos B = 2R sin B cos B = R sin 2B.

Likewise, |DE| = c cos C = R sin 2C. Thus,

|DE| + |DF | = R(sin 2B + sin 2C). (†)

Since 0◦ < A, B, C < 90◦, by the sum-to-product formula,

|BC| − (|DE| + |DF |) = R[2 sin A − (sin 2B + sin 2C]= R[2 sin A − 2 sin (B + C) cos (B − C)]= 2R sin A[1 − cos (B − C)] ≥ 0,

as desired. (This can also be proven by the law of cosines.)

A

B CD

E

F

H

Figure 5.13.


Because both ABDE and ACDF are cyclic, � BDF = � CDE = � CAB.Thus, by the Lemma, we have

2([BFC] + [BEC])= |DF | · |BC| · sin � BDF + |DE| · |BC| · sin � EDC

= |BC|(|DE| + |DF |) sin A ≥ (|DE| + |DF |)2 sin A.

By equation (†), the last inequality is equivalent to

R2(sin 2B + sin 2C)2 sin A ≤ 2[BFC] + 2[BEC].

Likewise, we have

R2(sin 2C + sin 2A)2 sin B ≤ 2[CDA] + 2[CFA]

andR2(sin 2A + sin 2B)2 sin C ≤ 2[AEB] + 2[ADB].

Adding the last three inequalities yields the desired result. In view of theLemma, it is also clear that equality holds if and only if triangle ABC isequilateral.

49. [Bulgaria 1998] On the sides of a nonobtuse triangle ABC are constructedexternally a square P4, a regular m-sided polygon Pm, and a regular n-sidedpolygon Pn. The centers of the square and the two polygons form an equilateraltriangle. Prove that m = n = 6, and find the angles of triangle ABC.

Solution: The angles are 90◦, 45◦, and 45◦. We prove the following lemma.

Lemma Let O be a point inside equilateral triangle XYZ. If

� YOZ = x, � ZOX = y, � XOY = z.

then|OX|

sin(x − 60◦)= |OY |

sin(y − 60◦)= |OZ|

sin(z − 60◦).

Proof: As shown in Figure 5.14, let R denote clockwise rotation of 60◦around the point Z, and let R(X) = X1 and R(O) = O1.


X

Y Z

O

O1

X1

Figure 5.14.

Then R(Y ) = X, and so triangle ZO1O is equilateral. Consequently, trianglesZO1X and ZOY are congruent, and so |O1X| = |OY |. Note that x +y +z =360◦. We have

� O1OX = � ZOX − � ZOO1 = � ZOX − 60◦ = y − 60◦,� XO1O = � XO1Z − � OO1Z = � YOZ − 60◦ = x − 60◦,� OXO1 = 180◦ − � O1OX − � XO1O = z − 60◦.

Applying the law of sines to triangle XOO1 establishes the desired result.

Now we prove our main result. Without loss of generality, suppose that P4, Pm,and Pn are on sides AB, BC, and CA, respectively (Figure 5.15). Let O bethe circumcenter of triangle ABC. Without loss of generality, we assume thatthe circumradius of triangle ABC is 1, so |OA| = |OB| = |OC| = 1. Let X,Y , and Z be the centers of P4, Pm, and Pn, respectively.

A B

C

X

YZ

O

Figure 5.15.

Because |OB| = |OC| and |YB| = |YC|, BOCY is a kite with OY as its

axis of symmetry. Thus, � BOY = � BOC2 = � A, and � OYB = 180◦/m. Let


α = 180◦m

. Applying the law of sines to triangle OBY , we obtain

|OY | = sin (A + α)

sin α.

Likewise, by setting � ZOC = 180◦n

= β, we have

|OX| = sin(C + 45◦)sin 45◦ = √

2 sin(C + 45◦) and |OZ| = sin(B + β)

sin β.

Note that O is inside triangle XYZ, because it is on the respective per-pendicular rays from X, Y , and Z toward sides AB, BC, and CA. Because� BOY = � A and � BOX = � C, we find that � XOY = � C +� A. Likewise,� YOZ = � A + � B and � ZOX = � B + � C. Applying the Lemma gives

|OY |sin(B + C − 60◦)

= |OZ|sin(C + A − 60◦)

= |OX|sin(A + B − 60◦)

,

or |OY |sin(A + 60◦)

= |OZ|sin(B + 60◦)

= |OX|sin(C + 60◦)

.

It follows that

sin(A + α) csc α

sin(A + 60◦)= sin(B + β) csc β

sin(B + 60◦)=

√2 sin(C + 45◦)sin(C + 60◦)

.

Because y = cot x is decreasing for x with 0◦ ≤ x ≤ 180◦, by the additionand subtraction formulas, the function

f (x) = sin(x − 15◦)sin x

= cos 15◦ − cot x sin 15◦

is increasing for x with 0◦ ≤ x ≤ 90◦. Consequently, because 0◦ ≤ C+60◦ ≤150◦ ( � C ≤ 90◦), it follows that

√2 sin(C + 45◦)sin(C + 60◦)

≤√

2 sin(90◦ + 45◦)sin(90◦ + 60◦)

= 2,

with equality if and only if C = 90◦. Therefore,

sin(A + α) csc α

sin(A + 60◦)= sin(B + β) csc β

sin(B + 60◦)≤ 2. (∗)

Because triangle ABC is nonobtuse, at least two of its angles are between 45◦and 90◦. Without loss of generality, we may assume that 45◦ ≤ B (≤ 90◦),


so sin(B + 60◦) > 0 and cot B ≤ 1. Then from the second inequality in therelation (∗), we have

sin(B + β) csc β ≤ 2 sin(B + 60◦),

orsin B cot β + cos B ≤ sin B + √

3 cos B,

by the addition and subtraction formulas. Dividing both sides of the aboveinequality by sin B yields

cot β ≤ 1 + (√

3 − 1) cot B ≤ 1 + √3 − 1 = √

3,

implying that β ≥ 30◦. But since n ≥ 6, β = 180◦/n ≤ 30◦. Thus all theequalities hold, and so � C = 90◦ and � A = � B = 45◦, as claimed.

50. [MOSP 2000] Let ABC be an acute triangle. Prove that(cos A

cos B

)2

+(

cos B

cos C

)2

+(

cos C

cos A

)2

+ 8 cos A cos B cos C ≥ 4.

Note: It is easier to rewrite the above inequality in terms of cos2 A, cos2 B,and cos2 C. By Introductory Problem 24(d), we have

4 − 8 cos A cos B cos C = 4(

cos2 A + cos2 B + cos2 C)

.

It suffices to prove(cos A

cos B

)2

+(

cos B

cos C

)2

+(

cos C

cos A

)2

≥ 4(


. (†)

We present three approaches.

First Solution: By the weighted arithmetic–geometric means inequality,we have

2

(cos A

cos B

)2

+(

cos B

cos C

)2

≥ 33

√cos4 A

cos2 B cos2 C

= 3 cos2 A3√

cos2 A cos2 B cos2 C

≥ 12 cos2 A,


by Introductory Problem 28(a).Adding the above inequality with its analogousforms and dividing both sides of the resulting inequality by 3, we obtaininequality (†).

Second Solution: Setting x = cos Bcos C

, y = cos Ccos A

, z = cos Acos B

in AdvancedProblem 42(a) yields(

cos A

cos B

)2

+(

cos B

cos C

)2

+(

cos C

cos A

)2

= x2 + y2 + z2

≥ 2(yz cos A + zx cos B + xy cos C)

= 2

[cos C cos A

cos B+ cos A cos B

cos C+ cos B cos C

cos A

].

However, setting

x =√

cos B cos C

cos A, y =

√cos A cos B

cos C, z =

√cos C cos A

cos B,

in Advanced Problem 42(a) again, we find that

2

[cos C cos A

cos B+ cos A cos B

cos C+ cos B cos C

cos A

]= 2(x2 + y2 + z2)

≥ 4(yz cos A + zx cos B + xy cos C)

= 4(


,

by noting that

yz cos A = cos A

√cos A cos B

cos C· cos C cos A

cos B= cos2 A

and its analogous forms for zx cos B and xy cos C.

Third Solution: The result follows from the following Lemma.

Lemma For positive real numbers a, b, c such that abc ≤ 1,

a

b+ b

c+ c

a≥ a + b + c.

Proof: Replacing a, b, c by ta, tb, tc with t = 1/3√

abc leaves the left-handside of the inequality unchanged and increases the value of the right-hand


side and results in the equality atbtct = abct3 = 1. Hence we may assumewithout loss of generality that abc = 1. Then there exist positive real numbersx, y, z such that a = x/y, b = z/x, c = y/z. The rearrangement inequalitygives

x3 + y3 + z3 ≥ x2z + y2x + z2y.

Thusa

b+ b

c+ c

a= x2

yz+ y2

zx+ z2

xy= x3 + y3 + z3

xyz

≥ x2z + y2x + z2y

xyz= x

y+ y

z+ z

x

= a + b + c,

as desired.

Now we prove our main result. Note that(4 cos2 A

) (4 cos2 B

) (4 cos2 C

)= (8 cos A cos B cos C)2 ≤ 1

by Introductory Problem 28(a). Setting a = 4 cos2 A, b = 4 cos2 B, c =4 cos2 C in the Lemma yields(

cos A

cos B

)2

+(

cos B

cos C

)2

+(

cos C

cos A

)2

= a

b+ b

c+ c

a≥ a + b + c

= 4(cos2 A + cos2 B + cos2 C),

establishing inequality (†).

51. For any real number x and any positive integer n, prove that∣∣∣∣∣n∑

k=1

sin kx

k

∣∣∣∣∣ ≤ 2√

π.

Solution: The solution is based on the following three Lemmas.

Lemma 1 Letnbe a positive integer, and leta1, a2, . . . , an andb1, b2, . . . , bn

be two sequences of real numbers. Then

n∑k=1

akbk = Snbn +n−1∑k=1

Sk(bk − bk+1),


where Sk = a1 + a2 + · · · + ak , for k = 1, 2, . . . , n.

Proof: Set S0 = 0. Then ak = Sk − Sk−1 for k = 1, 2, . . . , n, and so

n∑k=1

akbk =n∑

k=1

(Sk − Sk−1)bk =n∑

k=1

Skbk −n∑

k=1

Sk−1bk

= Snbn +n−1∑k=1

Skbk −n∑

k=2

Sk−1bk − S0b1

= Snbn +n−1∑k=1

Skbk −n−1∑k=1

Skbk+1

= Snbn +n−1∑k=1

Sk(bk − bk+1),

as desired.

Lemma 2 [Abel’s inequality] Letnbe a positive integer, and leta1, a2, . . . , an

and b1, b2, . . . , bn be two sequences of real numbers with b1 ≥ b2 ≥ · · · ≥bn ≥ 0. Then

mb1 ≤n∑

k=1

akbk ≤ Mb1,

where Sk = a1 + a2 + · · · + ak , for k = 1, 2, . . . , n, and M and m are themaximum and minimum, respectively, of { S1, S2, . . . , Sn }.Proof: Note that bn ≥ 0 and bk − bk+1 ≥ 0 for k = 1, 2, . . . , n− 1. Lemma1 gives

n∑k=1

akbk = Snbn +n−1∑k=1

Sk(bk − bk+1)

≤ Mbn + M

n−1∑k=1

(bk − bk+1) = Mb1.

establishing the second desired inequality. In exactly the same way, we canprove the first desired inequality.

Lemma 3 Let x be a real number that is not an even multiple of π , then∣∣∣∣∣n∑

k=m+1

sin kx

k

∣∣∣∣∣ ≤ 1

(m + 1)∣∣sin x

2

∣∣ ,


where m and n are positive integers with m < n.

Proof: For k = 1, 2, . . . , n−m, let ak = sin[(k+m)x] sin x2 and bk = 1

k+m.

Then by Lemma 2, we have

s

m + 1= sb1 ≤

n∑k=m+1

sin kx sin x2

k=

n−m∑k=1

akbk ≤ Sb1 = S

m + 1,

where Sk = a1 + a2 + · · · + ak , and S = max{S1, S2, . . . Sn} and s =min{S1, S2, . . . , Sn}. The product-to-sum formulas give

2ai = 2 sin[(i + m)x] sinx

2

= cos

(i + m − 1

2

)x − cos

(i + m + 1

2

)x,

and so

2Sk = 2a1 + 2a2 + · · · + 2ak = cos

(m + 1

2

)x − cos

(k + m + 1

2

)x.

It follows that −2 ≤ 2Sk ≤ 2 for k = 1, 2, . . . , n, and so −1 ≤ s ≤ S ≤ 1.Consequently,

− 1

m + 1= −b1 ≤ sb1 ≤

n∑k=m+1

sin kx sin x2

k≤ Sb1 ≤ b1 = 1

m + 1,

implying that ∣∣∣∣∣n∑

k=m+1

sin kx sin x2

k

∣∣∣∣∣ ≤ 1

m + 1,


Now we are ready to prove our main result. Because y = | sin x| is a periodicfunction with period π , we may assume that x is in the interval (0, π). (Notethat the desired result is trivial for x = 0.) For a fixed x with 0 < x < π , letm be the nonnegative integer such that

m ≤√

π

x< m + 1.

Thus ∣∣∣∣∣n∑

k=1

sin kx

k

∣∣∣∣∣ ≤∣∣∣∣∣

m∑k=1

sin kx

k

∣∣∣∣∣+∣∣∣∣∣

n∑k=m+1

sin kx

k

∣∣∣∣∣ .


Here we set the first summation on the right-hand side to be 0 if m = 0, andthe first summation taken from 1 to n and the second to be 0 if m ≥ n. Itsuffices to show that ∣∣∣∣∣

m∑k=1

sin kx

k

∣∣∣∣∣ ≤ √π (∗)

and ∣∣∣∣∣n∑

k=m+1

sin kx

k

∣∣∣∣∣ ≤ √π. (∗∗)

Because | sin x| < x and by the definition of m, it follows that∣∣∣∣∣m∑

k=1

sin kx

k

∣∣∣∣∣ ≤m∑

k=1

kx

k=

m∑k=1

x = mx ≤ √π,

establishing inequality (∗). On the other hand, by Lemma 3, we have∣∣∣∣∣n∑

k=m+1

sin kx

k

∣∣∣∣∣ ≤ 1

(m + 1)∣∣sin x

2

∣∣ .Note that y = sin x is concave for 0 < x < π

2 . Thus, the graph of y = sin x isabove the line connecting the points (0, 0) and

(π2 , 1)

on the interval(0, π

2

);

that is, sin x > 2xπ

. Hence for 0 < x < π , we have

sinx

2<

2 · x2

π= x

π.

It follows that∣∣∣∣∣n∑

k=m+1

sin kx

k

∣∣∣∣∣ ≤ 1

(m + 1)∣∣sin x

2

∣∣ ≤ 1

m + 1· x

π≤

√π

x· x

π= √

π,

establishing inequality (∗∗). Our proof is thus complete.

Glossary

Arithmetic–Geometric Means Inequality

If n is a positive integer and a1, a2, . . . , an are nonnegative real numbers, then

1

n

n∑i=1

ai ≥ (a1a2 · · · an)1/n,

with equality if and only if a1 = a2 = · · · = an. This inequality is a special case ofthe power mean inequality.

Arithmetic–Harmonic Means Inequality

If a1, a2, . . . , an are n positive numbers, then

1

n

n∑i=1

ai ≥ 11n

∑ni=1

1ai

,

with equality if and only if a1 = a2 = · · · = an. This inequality is a special case ofthe power mean inequality.


Binomial Coefficient (n

k

)= n!

k!(n − k)! ,

the coefficient of xk in the expansion of (x + 1)n.

Cauchy–Schwarz Inequality

For any real numbers a1, a2, . . . , an, and b1, b2, . . . , bn

(a21 + a2

2 + · · · + a2n)(b

21 + b2

2 + · · · + b2n)

≥ (a1b1 + a2b2 + · · · + anbn)2,

with equality if and only if ai and bi are proportional, i = 1, 2, . . . , n.

Ceva’s Theorem and Its Trigonometric Form

Let AD, BE, CF be three cevians of triangle ABC. The following are equivalent:

(i) AD, BE, CF are concurrent;

(ii)|AF ||FB| · |BD|

|DC| · |CE||EA| = 1;

(iii)sin � ABE

sin � EBC· sin � BCF

sin � FCA· sin � CAD

sin � DAB= 1.

Cevian

A cevian of a triangle is any segment joining a vertex to a point on the opposite side.

Chebyshev’s Inequality

1. Let x1, x2, . . . , xn and y1, y2, . . . , yn be two sequences of real numbers suchthat x1 ≤ x2 ≤ · · · ≤ xn and y1 ≤ y2 ≤ · · · ≤ yn. Then

1

n(x1 + x2 + · · · + xn)(y1 + y2 + · · · + yn) ≤ x1y1 + x2y2 + · · · + xnyn.

2. Let x1, x2, . . . , xn and y1, y2, . . . , yn be two sequences of real numbers suchthat x1 ≥ x2 ≥ · · · ≥ xn and y1 ≥ y2 ≥ · · · ≥ yn. Then

1

n(x1 + x2 + · · · + xn)(y1 + y2 + · · · + yn) ≥ x1y1 + x2y2 + · · · + xnyn.

Glossary 201

Chebyshev Polynomials

Let {Tn(x)}∞n=0 be the sequence of polynomials such that T0(x) = 1, T1(x) = x,and Ti+1 = 2xTi(x) − Ti−1(x) for all positive integers i. The polynomial Tn(x) iscalled the nth Chebyshev polynomial.

Circumcenter

The center of the circumscribed circle or sphere.

Circumcircle

A circumscribed circle.

Convexity

A function f (x) is concave up (down) on [a, b] ⊆ R if f (x) lies under (over) theline connecting (a1, f (a1)) and (b1, f (b1)) for all

a ≤ a1 < x < b1 ≤ b.

Concave up and down functions are also called convex and concave, respectively.If f is concave up on an interval [a, b] and λ1, λ2, . . ., λn are nonnegative numbers

with sum equal to 1, then

λ1f (x1) + λ2f (x2) + · · · + λnf (xn) ≥ f (λ1x1 + λ2x2 + · · · + λnxn)

for any x1, x2, . . . , xn in the interval [a, b]. If the function is concave down, theinequality is reversed. This is Jensen’s inequality.

Cyclic Sum

Let n be a positive integer. Given a function f of n variables, define the cyclic sumof variables (x1, x2, . . . , xn) as∑

cyc

f (x1, x2, . . . , xn) = f (x1, x2, . . . , xn) + f (x2, x3, . . . , xn, x1)

+ · · · + f (xn, x1, x2, . . . , xn−1).

De Moivre’s Formula

For any angle α and for any integer n,

(cos α + i sin α)n = cos nα + i sin nα.


From this formula, we can easily derive the expansion formulas of sin nα and cos nα

in terms of sin α and cos α.

Euler’s Formula (in Plane Geometry)

Let O and I be the circumcenter and incenter, respectively, of a triangle with cir-cumradius R and inradius r . Then

|OI |2 = R2 − 2rR.

Excircles, or Escribed Circles

Given a triangle ABC, there are four circles tangent to the lines AB, BC, CA. One isthe inscribed circle, which lies in the interior of the triangle. One lies on the oppositeside of line BC from A, and is called the excircle (escribed circle) opposite A, andsimilarly for the other two sides. The excenter opposite A is the center of the excircleopposite A; it lies on the internal angle bisector of A and the external angle bisectorsof B and C.

Excenters

See Excircles.

Extended Law of Sines

In a triangle ABC with circumradius equal to R,

|BC|sin A

= |CA|sin B

= |AB|sin C

= 2R.

Gauss’s Lemma

Letp(x) = anx

n + an−1xn−1 + · · · + aax + a0

be a polynomial with integer coefficients. All the rational roots (if there are any) ofp(x) can be written in the reduced form m

n, where m and n are divisors of a0 and an,

respectively.

Gergonne Point

If the incircle of triangle ABC touches sides AB, BC, and CA at F, D, and E, thenlines AD, BE, and CF are concurrent, and the point of concurrency is called theGergonne point of the triangle.

Glossary 203

Heron’s Formula

The area of a triangle ABC with sides a, b, c is equal to

[ABC] = √s(s − a)(s − b)(s − c),

where s = (a + b + c)/2 is the semiperimeter of the triangle.

Homothety

A homothety (central similarity) is a transformation that fixes one point O (its center)and maps each point P to a point P ′ for which O, P, P ′ are collinear and the ratio|OP | : |OP ′| = k is constant (k can be either positive or negative); k is called themagnitude of the homothety.

Homothetic Triangles

Two triangles ABC and DEF are homothetic if they have parallel sides. Supposethat AB ‖ DE, BC ‖ EF , and CA ‖ FD. Then lines AD, BE, and CF concurat a point X, as given by a special case of Desargues’s theorem. Furthermore, somehomothety centered at X maps triangle ABC onto triangle DEF .

Incenter

The center of an inscribed circle.

Incircle

An inscribed circle.

Jensen’s Inequality

See Convexity.

Kite

A quadrilateral with its sides forming two pairs of congruent adjacent sides. A kiteis symmetric with one of its diagonals. (If it is symmetric with both diagonals, itbecomes a rhombus.) The two diagonals of a kite are perpendicular to each other.For example, if ABCD is a quadrilateral with |AB| = |AD| and |CB| = |CD|,then ABCD is a kite, and it is symmetric with respect to the diagonal AC.


Lagrange’s Interpolation Formula

Let x0, x1, . . . , xn be distinct real numbers, and let y0, y1, . . . , yn be arbitrary realnumbers. Then there exists a unique polynomial P(x) of degree at most n such thatP(xi) = yi , i = 0, 1, . . . , n. This polynomial is given by

P(x) =n∑

i=0

yi(x − x0) · · · (x − xi−1)(x − xi+1) · · · (x − xn)

(xi − x0) · · · (xi − xi−1)(xi − xi+1) · · · (xi − xn).

Law of Cosines

In a triangle ABC,

|CA|2 = |AB|2 + |BC|2 − 2|AB| · |BC| cos � ABC,

and analogous equations hold for |AB|2 and |BC|2.

Median formula

This is also called the length of the median formula. Let AM be a median in triangleABC. Then

|AM|2 = 2|AB|2 + 2|AC|2 − |BC|24

.

Minimal Polynomial

We call a polynomial p(x) with integer coefficients irreducible if p(x) cannot bewritten as a product of two polynomials with integer coefficients neither of whichis a constant. Suppose that the number α is a root of a polynomial q(x) with integercoefficients.Among all polynomials with integer coefficients with leading coefficient1 (i.e., monic polynomials with integer coefficients) that have α as a root, there isone of smallest degree. This polynomial is the minimal polynomial of α. Let p(x)

denote this polynomial. Then p(x) is irreducible, and for any other polynomial q(x)

with integer coefficients such that q(α) = 0, the polynomial p(x) divides q(x); thatis, q(x) = p(x)h(x) for some polynomial h(x) with integer coefficients.

Orthocenter of a Triangle

The point of intersection of the altitudes.

Glossary 205

Periodic Function

A function f (x) is periodic with period T > 0 if T is the smallest positive realnumber for which

f (x + T ) = f (x)

for all x.

Pigeonhole Principle

If n objects are distributed among k < n boxes, some box contains at least twoobjects.

Power Mean Inequality

Let a1, a2, . . . , an be any positive numbers for which a1 + a2 + · · · + an = 1. Forpositive numbers x1, x2, . . . , xn we define

M−∞ = min{x1, x2, . . . , xk},M∞ = max{x1, x2, . . . , xk},M0 = x

a11 x

a22 · · · xan

n ,

Mt = (a1xt1 + a2x

t2 + · · · + akx

tk

)1/t,

where t is a nonzero real number. Then

M−∞ ≤ Ms ≤ Mt ≤ M∞for s ≤ t .

Rearrangement Inequality

Let a1 ≤ a2 ≤ · · · ≤ an; b1 ≤ b2 ≤ · · · ≤ bn be real numbers, and let c1, c2, . . . , cn

be any permutations of b1 ≤ b2 ≤ · · · ≤ bn. Then

a1bn + a2bn−1 + · · · + anb1 ≤ a1c1 + a2c2 + · · · + ancn

≤ a1b1 + a2b2 + · · · + anbn,

with equality if and only if a1 = a2 = · · · = an or b1 = b2 = · · · = bn.

Root Mean Square–Arithmetic Mean Inequality

For positive numbers x1, x2, . . . , xn,√x2

1 + x22 + · · · + x2

k

n≥ x1 + x2 + · · · + xk

n.


The inequality is a special case of the power mean inequality.

Schur’s Inequality

Let x, y, z be nonnegative real numbers. Then for any r > 0,

xr(x − y)(x − z) + yr(y − z)(y − x) + zr(z − x)(z − y) ≥ 0.

Equality holds if and only if x = y = z or if two of x, y, z are equal and the third isequal to 0.

The proof of the inequality is rather simple. Because the inequality is symmetricin the three variables, we may assume without loss of generality that x ≥ y ≥ z.Then the given inequality may be rewritten as

(x − y)[xr(x − z) − yr(y − z)

]+ zr(x − z)(y − z) ≥ 0,

and every term on the left-hand side is clearly nonnegative. The first term is positiveif x > y, so equality requires x = y, as well as zr(x − z)(y − z) = 0, which giveseither x = y = z or z = 0.

Sector

The region enclosed by a circle and two radii of the circle.

Stewart’s Theorem

In a triangle ABC with cevian AD, write a = |BC|, b = |CA|, c = |AB|, m =|BD|, n = |DC|, and d = |AD|. Then

d2a + man = c2n + b2m.

This formula can be used to express the lengths of the altitudes and angle bisectorsof a triangle in terms of its side lengths.

Trigonometric Identities

sin2 a + cos2 a = 1,

1 + cot2 a = csc2 a,

tan2 x + 1 = sec2 x.

Glossary 207

Addition and Subtraction Formulas:

sin(a ± b) = sin a cos b ± cos a sin b,

cos(a ± b) = cos a cos b ∓ sin a sin b,

tan(a ± b) = tan a ± tan b

1 ∓ tan a tan b,

cot(a ± b) = cot a cot b ∓ 1

cot a ± cot b.

Double-Angle Formulas:

sin 2a = 2 sin a cos a = 2 tan a

1 + tan2 a,

cos 2a = 2 cos2 a − 1 = 1 − 2 sin2 a = 1 − tan2 a

1 + tan2 a,

tan 2a = 2 tan a

1 − tan2 a;

cot 2a = cot2 a − 1

2 cot a.

Triple-Angle Formulas:

sin 3a = 3 sin a − 4 sin3 a,

cos 3a = 4 cos3 a − 3 cos a,

tan 3a = 3 tan a − tan3 a

1 − 3 tan2 a.

Half-Angle Formulas:

sin2 a

2= 1 − cos a

2,

cos2 a

2= 1 + cos a

2,

tana

2= 1 − cos a

sin a= sin a

1 + cos a,

cota

2= 1 + cos a

sin a= sin a

1 − cos a.


Sum-to-Product Formulas:

sin a + sin b = 2 sina + b

2cos

a − b

2,

cos a + cos b = 2 cosa + b

2cos

a − b

2,

tan a + tan b = sin(a + b)

cos a cos b.

Difference-to-Product Formulas:

sin a − sin b = 2 sina − b

2cos

a + b

2,

cos a − cos b = −2 sina − b

2sin

a + b

2,

tan a − tan b = sin(a − b)

cos a cos b.

Product-to-Sum Formulas:

2 sin a cos b = sin(a + b) + sin(a − b),

2 cos a cos b = cos(a + b) + cos(a − b),

2 sin a sin b = − cos(a + b) + cos(a − b).

Expansion Formulas

sin nα =(

n

1

)cosn−1 α sin α −

(n

3


+(

n

5

)cosn−5 α sin5 α − · · · ,

cos nα =(

n

0

)cosn α −

(n

2


+(

n

4

)cosn−4 α sin4 α − · · · .

Viète’s Theorem

Let x1, x2, . . . , xn be the roots of polynomial

P(x) = anxn + an−1x

n−1 + · · · + a1x + a0,

where an �= 0 and a0, a1, . . . , an ∈ C. Let sk be the sum of the products of the xi

taken k at a time. Thensk = (−1)k

an−k

an

;

Glossary 209

that is,

x1 + x2 + · · · + xn = −an−1

an

;

x1x2 + · · · + xixj + xn−1xn = an−2

an

;...

x1x2 · · · xn = (−1)na0

an

.

Further Reading

1. Andreescu, T.; Feng, Z., 101 Problems in Algebra from the Training of theUSA IMO Team, Australian Mathematics Trust, 2001.

2. Andreescu, T.; Feng, Z., 102 Combinatorial Problems from the Training of theUSA IMO Team, Birkhäuser, 2002.

3. Andreescu, T.; Feng, Z., USA and International Mathematical Olympiads2003, Mathematical Association of America, 2004.




7. Andreescu, T.; Feng, Z.; Lee, G.; Loh, P., Mathematical Olympiads: Problemsand Solutions from around the World, 2001–2002, Mathematical Associationof America, 2004.

8. Andreescu, T.; Feng, Z.; Lee, G., Mathematical Olympiads: Problems andSolutions from around the World, 2000–2001, Mathematical Association ofAmerica, 2003.


9. Andreescu, T.; Feng, Z., Mathematical Olympiads: Problems and Solutionsfrom around the World, 1999–2000, Mathematical Association of America,2002.

10. Andreescu, T.; Feng, Z., Mathematical Olympiads: Problems and Solutionsfrom around the World, 1998–1999, Mathematical Association of America,2000.

11. Andreescu, T.; Kedlaya, K., Mathematical Contests 1997–1998: OlympiadProblems from around theWorld, with Solutions,American Mathematics Com-petitions, 1999.

12. Andreescu, T.; Kedlaya, K., Mathematical Contests 1996–1997: OlympiadProblems from around theWorld, with Solutions,American Mathematics Com-petitions, 1998.

13. Andreescu, T.; Kedlaya, K.; Zeitz, P., Mathematical Contests 1995–1996:Olympiad Problems from around the World, with Solutions, American Math-ematics Competitions, 1997.

14. Andreescu, T.; Enescu, B., Mathematical Olympiad Treasures, Birkhäuser,2003.

15. Andreescu, T.; Gelca, R., Mathematical Olympiad Challenges, Birkhäuser,2000.

16. Andreescu, T.; Andrica, D., 360 Problems for Mathematical Contests, GILPublishing House, 2003.

17. Andreescu, T.; Andrica, D., Complex Numbers from A to Z, Birkhäuser, 2004.

18. Beckenbach, E. F.; Bellman, R., An Introduction to Inequalities, New Mathe-matical Library, Vol. 3, Mathematical Association of America, 1961.

19. Coxeter, H. S. M.; Greitzer, S. L., Geometry Revisited, New MathematicalLibrary, Vol. 19, Mathematical Association of America, 1967.

20. Coxeter, H. S. M., Non-Euclidean Geometry, The Mathematical Associationof America, 1998.

21. Doob, M., The Canadian Mathematical Olympiad 1969–1993, University ofToronto Press, 1993.

22. Engel,A., Problem-Solving Strategies, Problem Books in Mathematics, Spring-er, 1998.

Further Reading 213

23. Fomin, D.; Kirichenko, A., Leningrad Mathematical Olympiads 1987–1991,MathPro Press, 1994.

24. Fomin, D.; Genkin, S.; Itenberg, I., Mathematical Circles, American Mathe-matical Society, 1996.

25. Graham, R. L.; Knuth, D. E.; Patashnik, O., Concrete Mathematics, Addison-Wesley, 1989.

26. Gillman, R., A Friendly Mathematics Competition, The Mathematical Asso-ciation of America, 2003.

27. Greitzer, S. L., International Mathematical Olympiads, 1959–1977, New Math-ematical Library, Vol. 27, Mathematical Association of America, 1978.

28. Holton, D., Let’s Solve Some Math Problems, A Canadian Mathematics Com-petition Publication, 1993.

29. Kazarinoff, N. D., Geometric Inequalities, New Mathematical Library, Vol. 4,Random House, 1961.

30. Kedlaya, K; Poonen, B.; Vakil, R., The William Lowell Putnam MathematicalCompetition 1985–2000, The Mathematical Association of America, 2002.

31. Klamkin, M., International Mathematical Olympiads, 1978–1985, New Math-ematical Library, Vol. 31, Mathematical Association of America, 1986.

32. Klamkin, M., USA Mathematical Olympiads, 1972–1986, New MathematicalLibrary, Vol. 33, Mathematical Association of America, 1988.

33. Kürschák, J., Hungarian Problem Book, volumes I & II, New MathematicalLibrary, Vols. 11 & 12, Mathematical Association of America, 1967.

34. Kuczma, M., 144 Problems of the Austrian–Polish Mathematics Competition1978–1993, The Academic Distribution Center, 1994.

35. Kuczma, M., International Mathematical Olympiads 1986–1999, Mathemat-ical Association of America, 2003.

36. Larson, L. C., Problem-Solving Through Problems, Springer-Verlag, 1983.

37. Lausch, H. The Asian Pacific Mathematics Olympiad 1989–1993, AustralianMathematics Trust, 1994.

38. Liu, A., Chinese Mathematics Competitions and Olympiads 1981–1993, Aus-tralian Mathematics Trust, 1998.


39. Liu, A., Hungarian Problem Book III, New Mathematical Library, Vol. 42,Mathematical Association of America, 2001.

40. Lozansky, E.; Rousseau, C. Winning Solutions, Springer, 1996.

41. Mitrinovic, D. S.; Pecaric, J. E.; Volonec, V. Recent Advances in GeometricInequalities, Kluwer Academic Publisher, 1989.

42. Savchev, S.; Andreescu, T. Mathematical Miniatures, Anneli Lax New Math-ematical Library, Vol. 43, Mathematical Association of America, 2002.

43. Sharygin, I. F., Problems in Plane Geometry, Mir, Moscow, 1988.

44. Sharygin, I. F., Problems in Solid Geometry, Mir, Moscow, 1986.

45. Shklarsky, D. O; Chentzov, N. N;Yaglom, I. M., The USSR Olympiad ProblemBook, Freeman, 1962.

46. Slinko,A., USSR Mathematical Olympiads 1989–1992,Australian Mathemat-ics Trust, 1997.

47. Szekely, G. J., Contests in Higher Mathematics, Springer-Verlag, 1996.

48. Taylor, P. J., Tournament of Towns 1980–1984, Australian Mathematics Trust,1993.



51. Taylor, P. J.; Storozhev,A., Tournament of Towns 1993–1997,Australian Math-ematics Trust, 1998.

52. Yaglom, I. M., Geometric Transformations, New Mathematical Library, Vol.8, Random House, 1962.

53. Yaglom, I. M., Geometric Transformations II, New Mathematical Library, Vol.21, Random House, 1968.

54. Yaglom, I. M., Geometric Transformations III, New Mathematical Library,Vol. 24, Random House, 1973.

103 trigonometry problems

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